Mixing
iterations on a calculator
$begingroup$
Ok, first off I am not a mathematician, so I apologize if terms are misused, I will try to get my questions across with plain English so I don't stumble on using a term that may mean something specific to a mathematician.
Starting with "iterations" (it may mean something specific - but I mean it in general)
For instance - here is the scenario hat made me want to do something I couldn't figure out how to do - but I am sure there is a way to express it in a formula to get my answer.
Proposition - Put a quarter a day in the jar - have $9125 in a year...
Clearly this assertion was wrong, and simple math (.25*365) tells anyone that you'll have $91.25 in a year.
But then I wanted to see what would happen in you started at .25 and "added a quarter to the last amount added every day...what would you get?
> day 1 = add .25 (total=.25)
> day 2 = add .25 + .25 (total=.75)
> day 3 =
> add .5 + .25 (total=1.5)
> day 4 = add .75 + .25 (total=2.25)
> etc... to 365 days
How would you express this on a calculator (normal generally available software calculator or spreadsheet)
Like wise (second question) - how would you figure doubling an amount for x iterations?
Thanks
sequences-and-series
asked Jan 16 at 16:55
jpmyobjpmyob
1061
$endgroup$
add a comment |
$begingroup$
Ok, first off I am not a mathematician, so I apologize if terms are misused, I will try to get my questions across with plain English so I don't stumble on using a term that may mean something specific to a mathematician.
Starting with "iterations" (it may mean something specific - but I mean it in general)
For instance - here is the scenario hat made me want to do something I couldn't figure out how to do - but I am sure there is a way to express it in a formula to get my answer.
Proposition - Put a quarter a day in the jar - have $9125 in a year...
Clearly this assertion was wrong, and simple math (.25*365) tells anyone that you'll have $91.25 in a year.
But then I wanted to see what would happen in you started at .25 and "added a quarter to the last amount added every day...what would you get?
> day 1 = add .25 (total=.25)
> day 2 = add .25 + .25 (total=.75)
> day 3 =
> add .5 + .25 (total=1.5)
> day 4 = add .75 + .25 (total=2.25)
> etc... to 365 days
How would you express this on a calculator (normal generally available software calculator or spreadsheet)
Like wise (second question) - how would you figure doubling an amount for x iterations?
Thanks
sequences-and-series
asked Jan 16 at 16:55
jpmyobjpmyob
1061
$endgroup$
1
$begingroup$
The original can be phrased as a recursive sequence $a_0 = 0, a_n = a_{n-1}+0.25$ to which the closed form as you should clearly be able to see is $a_n = 0.25n$. The second question you pose is a recursive sequence of the form $a_0 =0, a_n = a_{n-1}+0.25n$ to which the closed form is $frac{1}{4}times frac{n(n+1)}{2}$, that is a quarter of the $n$'th triangle number. You would have $a_n = 16698.75$. For information on how to solve this and other such things by hand, look up linear recurrence equations.
$endgroup$
– JMoravitz
Jan 16 at 17:02
1
$begingroup$
In mathmatics this is called a "recursive definition" and it is define as $b_{n+1} = b_n + .25$ and our running total is $t_{n+1} = t_n + b_{n+1}$. In computer programing it can be in the form of a function that calls itself. function daily_add(n){if (n==0) return 0; else return daily_add(n-1) + .25;}
$endgroup$
– fleablood
Jan 16 at 17:04
add a comment |
$begingroup$
Ok, first off I am not a mathematician, so I apologize if terms are misused, I will try to get my questions across with plain English so I don't stumble on using a term that may mean something specific to a mathematician.
Starting with "iterations" (it may mean something specific - but I mean it in general)
For instance - here is the scenario hat made me want to do something I couldn't figure out how to do - but I am sure there is a way to express it in a formula to get my answer.
Proposition - Put a quarter a day in the jar - have $9125 in a year...
Clearly this assertion was wrong, and simple math (.25*365) tells anyone that you'll have $91.25 in a year.
But then I wanted to see what would happen in you started at .25 and "added a quarter to the last amount added every day...what would you get?
> day 1 = add .25 (total=.25)
> day 2 = add .25 + .25 (total=.75)
> day 3 =
> add .5 + .25 (total=1.5)
> day 4 = add .75 + .25 (total=2.25)
> etc... to 365 days
How would you express this on a calculator (normal generally available software calculator or spreadsheet)
Like wise (second question) - how would you figure doubling an amount for x iterations?
Thanks
sequences-and-series
asked Jan 16 at 16:55
jpmyobjpmyob
1061
$endgroup$
Ok, first off I am not a mathematician, so I apologize if terms are misused, I will try to get my questions across with plain English so I don't stumble on using a term that may mean something specific to a mathematician.
Starting with "iterations" (it may mean something specific - but I mean it in general)
For instance - here is the scenario hat made me want to do something I couldn't figure out how to do - but I am sure there is a way to express it in a formula to get my answer.
Proposition - Put a quarter a day in the jar - have $9125 in a year...
Clearly this assertion was wrong, and simple math (.25*365) tells anyone that you'll have $91.25 in a year.
But then I wanted to see what would happen in you started at .25 and "added a quarter to the last amount added every day...what would you get?
> day 1 = add .25 (total=.25)
> day 2 = add .25 + .25 (total=.75)
> day 3 =
> add .5 + .25 (total=1.5)
> day 4 = add .75 + .25 (total=2.25)
> etc... to 365 days
How would you express this on a calculator (normal generally available software calculator or spreadsheet)
Like wise (second question) - how would you figure doubling an amount for x iterations?
Thanks
sequences-and-series
sequences-and-series
asked Jan 16 at 16:55
jpmyobjpmyob
1061
asked Jan 16 at 16:55
jpmyobjpmyob
1061
asked Jan 16 at 16:55
jpmyobjpmyob
1061
asked Jan 16 at 16:55
jpmyobjpmyob
1061
asked Jan 16 at 16:55
jpmyobjpmyob
1061
1061
1
$begingroup$
The original can be phrased as a recursive sequence $a_0 = 0, a_n = a_{n-1}+0.25$ to which the closed form as you should clearly be able to see is $a_n = 0.25n$. The second question you pose is a recursive sequence of the form $a_0 =0, a_n = a_{n-1}+0.25n$ to which the closed form is $frac{1}{4}times frac{n(n+1)}{2}$, that is a quarter of the $n$'th triangle number. You would have $a_n = 16698.75$. For information on how to solve this and other such things by hand, look up linear recurrence equations.
$endgroup$
– JMoravitz
Jan 16 at 17:02
1
$begingroup$
In mathmatics this is called a "recursive definition" and it is define as $b_{n+1} = b_n + .25$ and our running total is $t_{n+1} = t_n + b_{n+1}$. In computer programing it can be in the form of a function that calls itself. function daily_add(n){if (n==0) return 0; else return daily_add(n-1) + .25;}
$endgroup$
– fleablood
Jan 16 at 17:04
add a comment |
1
$begingroup$
The original can be phrased as a recursive sequence $a_0 = 0, a_n = a_{n-1}+0.25$ to which the closed form as you should clearly be able to see is $a_n = 0.25n$. The second question you pose is a recursive sequence of the form $a_0 =0, a_n = a_{n-1}+0.25n$ to which the closed form is $frac{1}{4}times frac{n(n+1)}{2}$, that is a quarter of the $n$'th triangle number. You would have $a_n = 16698.75$. For information on how to solve this and other such things by hand, look up linear recurrence equations.
$endgroup$
– JMoravitz
Jan 16 at 17:02
1
$begingroup$
In mathmatics this is called a "recursive definition" and it is define as $b_{n+1} = b_n + .25$ and our running total is $t_{n+1} = t_n + b_{n+1}$. In computer programing it can be in the form of a function that calls itself. function daily_add(n){if (n==0) return 0; else return daily_add(n-1) + .25;}
$endgroup$
– fleablood
Jan 16 at 17:04
1
1
$begingroup$
The original can be phrased as a recursive sequence $a_0 = 0, a_n = a_{n-1}+0.25$ to which the closed form as you should clearly be able to see is $a_n = 0.25n$. The second question you pose is a recursive sequence of the form $a_0 =0, a_n = a_{n-1}+0.25n$ to which the closed form is $frac{1}{4}times frac{n(n+1)}{2}$, that is a quarter of the $n$'th triangle number. You would have $a_n = 16698.75$. For information on how to solve this and other such things by hand, look up linear recurrence equations.
$endgroup$
– JMoravitz
Jan 16 at 17:02
$begingroup$
The original can be phrased as a recursive sequence $a_0 = 0, a_n = a_{n-1}+0.25$ to which the closed form as you should clearly be able to see is $a_n = 0.25n$. The second question you pose is a recursive sequence of the form $a_0 =0, a_n = a_{n-1}+0.25n$ to which the closed form is $frac{1}{4}times frac{n(n+1)}{2}$, that is a quarter of the $n$'th triangle number. You would have $a_n = 16698.75$. For information on how to solve this and other such things by hand, look up linear recurrence equations.
$endgroup$
– JMoravitz
Jan 16 at 17:02
1
1
$begingroup$
In mathmatics this is called a "recursive definition" and it is define as $b_{n+1} = b_n + .25$ and our running total is $t_{n+1} = t_n + b_{n+1}$. In computer programing it can be in the form of a function that calls itself. function daily_add(n){if (n==0) return 0; else return daily_add(n-1) + .25;}
$endgroup$
– fleablood
Jan 16 at 17:04
$begingroup$
In mathmatics this is called a "recursive definition" and it is define as $b_{n+1} = b_n + .25$ and our running total is $t_{n+1} = t_n + b_{n+1}$. In computer programing it can be in the form of a function that calls itself. function daily_add(n){if (n==0) return 0; else return daily_add(n-1) + .25;}
$endgroup$
– fleablood
Jan 16 at 17:04
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $a_{n}$ denote the amount that you are adding to the jar on day $n$. So $$a_{1}=.25, a_{2}=.5, a_{3} = .75,..., a_{n}=.25n$$
Then the amount that you have in the jar at day $n$ is your starting amount, which in this case is $0$, plus all of the amounts you've added up to this point. That is, the amount you have at day $n$ is
$$sum_{k=1}^{n}a_{k}=sum_{k=1}^{n}.25k=.25sum_{k=1}^{n}k=.25frac{k(k+1)}{2}=frac{k(k+1)}{8}$$
Here I used the well-known identity
$$sum_{k=1}^{n}k=frac{k(k+1)}{2}$$
answered Jan 16 at 17:01
pwerthpwerth
3,243417
$endgroup$
add a comment |
$begingroup$
You asked how to perform your calculation in a spreadsheet. Here is one way to do it. I am assuming you are using Excel or something similar.
- Enter 0.25 in call A1.
- Enter =A1+0.25 in call A2.
- Enter =A1 in cell B1.
- Enter =B1+A2 in cell B2.
- Select and copy cells A2 and B2.
- Paste into the rectangular region with upper left hand corner cell A3 and lower right-hand corner B365.
When you are done, column A will contain the amount to be added each day, and column B will contain the total amount in your bank on each day, so the total amount in your bank on day 365 will appear in cell B365.
answered Jan 16 at 21:05
awkwardawkward
6,14511022
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Let $a_{n}$ denote the amount that you are adding to the jar on day $n$. So $$a_{1}=.25, a_{2}=.5, a_{3} = .75,..., a_{n}=.25n$$
Then the amount that you have in the jar at day $n$ is your starting amount, which in this case is $0$, plus all of the amounts you've added up to this point. That is, the amount you have at day $n$ is
$$sum_{k=1}^{n}a_{k}=sum_{k=1}^{n}.25k=.25sum_{k=1}^{n}k=.25frac{k(k+1)}{2}=frac{k(k+1)}{8}$$
Here I used the well-known identity
$$sum_{k=1}^{n}k=frac{k(k+1)}{2}$$
answered Jan 16 at 17:01
pwerthpwerth
3,243417
$endgroup$
add a comment |
$begingroup$
Let $a_{n}$ denote the amount that you are adding to the jar on day $n$. So $$a_{1}=.25, a_{2}=.5, a_{3} = .75,..., a_{n}=.25n$$
Then the amount that you have in the jar at day $n$ is your starting amount, which in this case is $0$, plus all of the amounts you've added up to this point. That is, the amount you have at day $n$ is
$$sum_{k=1}^{n}a_{k}=sum_{k=1}^{n}.25k=.25sum_{k=1}^{n}k=.25frac{k(k+1)}{2}=frac{k(k+1)}{8}$$
Here I used the well-known identity
$$sum_{k=1}^{n}k=frac{k(k+1)}{2}$$
answered Jan 16 at 17:01
pwerthpwerth
3,243417
$endgroup$
add a comment |
$begingroup$
Let $a_{n}$ denote the amount that you are adding to the jar on day $n$. So $$a_{1}=.25, a_{2}=.5, a_{3} = .75,..., a_{n}=.25n$$
Then the amount that you have in the jar at day $n$ is your starting amount, which in this case is $0$, plus all of the amounts you've added up to this point. That is, the amount you have at day $n$ is
$$sum_{k=1}^{n}a_{k}=sum_{k=1}^{n}.25k=.25sum_{k=1}^{n}k=.25frac{k(k+1)}{2}=frac{k(k+1)}{8}$$
Here I used the well-known identity
$$sum_{k=1}^{n}k=frac{k(k+1)}{2}$$
answered Jan 16 at 17:01
pwerthpwerth
3,243417
$endgroup$
Let $a_{n}$ denote the amount that you are adding to the jar on day $n$. So $$a_{1}=.25, a_{2}=.5, a_{3} = .75,..., a_{n}=.25n$$
Then the amount that you have in the jar at day $n$ is your starting amount, which in this case is $0$, plus all of the amounts you've added up to this point. That is, the amount you have at day $n$ is
$$sum_{k=1}^{n}a_{k}=sum_{k=1}^{n}.25k=.25sum_{k=1}^{n}k=.25frac{k(k+1)}{2}=frac{k(k+1)}{8}$$
Here I used the well-known identity
$$sum_{k=1}^{n}k=frac{k(k+1)}{2}$$
answered Jan 16 at 17:01
pwerthpwerth
3,243417
answered Jan 16 at 17:01
pwerthpwerth
3,243417
answered Jan 16 at 17:01
pwerthpwerth
3,243417
answered Jan 16 at 17:01
pwerthpwerth
3,243417
3,243417
add a comment |
add a comment |
$begingroup$
You asked how to perform your calculation in a spreadsheet. Here is one way to do it. I am assuming you are using Excel or something similar.
- Enter 0.25 in call A1.
- Enter =A1+0.25 in call A2.
- Enter =A1 in cell B1.
- Enter =B1+A2 in cell B2.
- Select and copy cells A2 and B2.
- Paste into the rectangular region with upper left hand corner cell A3 and lower right-hand corner B365.
When you are done, column A will contain the amount to be added each day, and column B will contain the total amount in your bank on each day, so the total amount in your bank on day 365 will appear in cell B365.
answered Jan 16 at 21:05
awkwardawkward
6,14511022
$endgroup$
add a comment |
$begingroup$
You asked how to perform your calculation in a spreadsheet. Here is one way to do it. I am assuming you are using Excel or something similar.
- Enter 0.25 in call A1.
- Enter =A1+0.25 in call A2.
- Enter =A1 in cell B1.
- Enter =B1+A2 in cell B2.
- Select and copy cells A2 and B2.
- Paste into the rectangular region with upper left hand corner cell A3 and lower right-hand corner B365.
When you are done, column A will contain the amount to be added each day, and column B will contain the total amount in your bank on each day, so the total amount in your bank on day 365 will appear in cell B365.
answered Jan 16 at 21:05
awkwardawkward
6,14511022
$endgroup$
add a comment |
$begingroup$
You asked how to perform your calculation in a spreadsheet. Here is one way to do it. I am assuming you are using Excel or something similar.
- Enter 0.25 in call A1.
- Enter =A1+0.25 in call A2.
- Enter =A1 in cell B1.
- Enter =B1+A2 in cell B2.
- Select and copy cells A2 and B2.
- Paste into the rectangular region with upper left hand corner cell A3 and lower right-hand corner B365.
When you are done, column A will contain the amount to be added each day, and column B will contain the total amount in your bank on each day, so the total amount in your bank on day 365 will appear in cell B365.
answered Jan 16 at 21:05
awkwardawkward
6,14511022
$endgroup$
You asked how to perform your calculation in a spreadsheet. Here is one way to do it. I am assuming you are using Excel or something similar.
- Enter 0.25 in call A1.
- Enter =A1+0.25 in call A2.
- Enter =A1 in cell B1.
- Enter =B1+A2 in cell B2.
- Select and copy cells A2 and B2.
- Paste into the rectangular region with upper left hand corner cell A3 and lower right-hand corner B365.
When you are done, column A will contain the amount to be added each day, and column B will contain the total amount in your bank on each day, so the total amount in your bank on day 365 will appear in cell B365.
answered Jan 16 at 21:05
awkwardawkward
6,14511022
answered Jan 16 at 21:05
awkwardawkward
6,14511022
answered Jan 16 at 21:05
awkwardawkward
6,14511022
answered Jan 16 at 21:05
awkwardawkward
6,14511022
6,14511022
add a comment |
add a comment |
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1
$begingroup$
The original can be phrased as a recursive sequence $a_0 = 0, a_n = a_{n-1}+0.25$ to which the closed form as you should clearly be able to see is $a_n = 0.25n$. The second question you pose is a recursive sequence of the form $a_0 =0, a_n = a_{n-1}+0.25n$ to which the closed form is $frac{1}{4}times frac{n(n+1)}{2}$, that is a quarter of the $n$'th triangle number. You would have $a_n = 16698.75$. For information on how to solve this and other such things by hand, look up linear recurrence equations.
$endgroup$
– JMoravitz
Jan 16 at 17:02
1
$begingroup$
In mathmatics this is called a "recursive definition" and it is define as $b_{n+1} = b_n + .25$ and our running total is $t_{n+1} = t_n + b_{n+1}$. In computer programing it can be in the form of a function that calls itself. function daily_add(n){if (n==0) return 0; else return daily_add(n-1) + .25;}
$endgroup$
– fleablood
Jan 16 at 17:04