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How do I notate this?
How to put this problem into equation?Continuous PercentageHow to notate a While loop?A Sample of Tritium-3 Decayed to 94.5% of its Original Amount After a Year. (a) What is the half-life of tritium-3?multiple loans multiple payers - how to snowball fairlyWorking out how to fill a gap with the least number of items between 2 sizes and using a 2nd item if required.Sum of infinitely many integrals over a bounded intervalHow to find possible values of sum of complex numbers with finite alphabetsProof for sum of digits of a number until sum is a single numberHow do you add, subtract, multiply, and divide infinite decimals?
$begingroup$
Let’s say that I have a whole pie, and I take $75%$ (or 3/4ths) of that pie. Then, I take $75%$ of the remaining quarter of the pie and add it to the original $75%$, I would have $93.7%$ of the pie. I’ll keep on taking turns taking $75%$ of the smaller portion and adding it to the larger portion. How would I write a mathematical notation to find the percentage of the larger portion after $x$ number of turns taking $75%$ of the smaller portion and adding it to the larger portion?
calculus summation summation-method
edited Mar 29 at 22:34
Xander Henderson
15.1k103556
asked Mar 29 at 22:28
T. OuT. Ou
6
$endgroup$
add a comment |
$begingroup$
Let’s say that I have a whole pie, and I take $75%$ (or 3/4ths) of that pie. Then, I take $75%$ of the remaining quarter of the pie and add it to the original $75%$, I would have $93.7%$ of the pie. I’ll keep on taking turns taking $75%$ of the smaller portion and adding it to the larger portion. How would I write a mathematical notation to find the percentage of the larger portion after $x$ number of turns taking $75%$ of the smaller portion and adding it to the larger portion?
calculus summation summation-method
edited Mar 29 at 22:34
Xander Henderson
15.1k103556
asked Mar 29 at 22:28
T. OuT. Ou
6
$endgroup$
1
$begingroup$
Start by writing out the steps. Look for a pattern.
$endgroup$
– Xander Henderson
Mar 29 at 22:33
1
$begingroup$
Think of the pie you don't take. Left behind there is $(0.25)^n$ of the original pie. Which means that you have taken $1-0.25^n$
$endgroup$
– Doug M
Mar 29 at 22:33
add a comment |
$begingroup$
Let’s say that I have a whole pie, and I take $75%$ (or 3/4ths) of that pie. Then, I take $75%$ of the remaining quarter of the pie and add it to the original $75%$, I would have $93.7%$ of the pie. I’ll keep on taking turns taking $75%$ of the smaller portion and adding it to the larger portion. How would I write a mathematical notation to find the percentage of the larger portion after $x$ number of turns taking $75%$ of the smaller portion and adding it to the larger portion?
calculus summation summation-method
edited Mar 29 at 22:34
Xander Henderson
15.1k103556
asked Mar 29 at 22:28
T. OuT. Ou
6
$endgroup$
Let’s say that I have a whole pie, and I take $75%$ (or 3/4ths) of that pie. Then, I take $75%$ of the remaining quarter of the pie and add it to the original $75%$, I would have $93.7%$ of the pie. I’ll keep on taking turns taking $75%$ of the smaller portion and adding it to the larger portion. How would I write a mathematical notation to find the percentage of the larger portion after $x$ number of turns taking $75%$ of the smaller portion and adding it to the larger portion?
calculus summation summation-method
calculus summation summation-method
edited Mar 29 at 22:34
Xander Henderson
15.1k103556
asked Mar 29 at 22:28
T. OuT. Ou
6
edited Mar 29 at 22:34
Xander Henderson
15.1k103556
asked Mar 29 at 22:28
T. OuT. Ou
6
edited Mar 29 at 22:34
Xander Henderson
15.1k103556
edited Mar 29 at 22:34
Xander Henderson
15.1k103556
edited Mar 29 at 22:34
Xander Henderson
15.1k103556
15.1k103556
asked Mar 29 at 22:28
T. OuT. Ou
6
asked Mar 29 at 22:28
T. OuT. Ou
6
asked Mar 29 at 22:28
T. OuT. Ou
6
6
1
$begingroup$
Start by writing out the steps. Look for a pattern.
$endgroup$
– Xander Henderson
Mar 29 at 22:33
1
$begingroup$
Think of the pie you don't take. Left behind there is $(0.25)^n$ of the original pie. Which means that you have taken $1-0.25^n$
$endgroup$
– Doug M
Mar 29 at 22:33
add a comment |
1
$begingroup$
Start by writing out the steps. Look for a pattern.
$endgroup$
– Xander Henderson
Mar 29 at 22:33
1
$begingroup$
Think of the pie you don't take. Left behind there is $(0.25)^n$ of the original pie. Which means that you have taken $1-0.25^n$
$endgroup$
– Doug M
Mar 29 at 22:33
1
1
$begingroup$
Start by writing out the steps. Look for a pattern.
$endgroup$
– Xander Henderson
Mar 29 at 22:33
$begingroup$
Start by writing out the steps. Look for a pattern.
$endgroup$
– Xander Henderson
Mar 29 at 22:33
1
1
$begingroup$
Think of the pie you don't take. Left behind there is $(0.25)^n$ of the original pie. Which means that you have taken $1-0.25^n$
$endgroup$
– Doug M
Mar 29 at 22:33
$begingroup$
Think of the pie you don't take. Left behind there is $(0.25)^n$ of the original pie. Which means that you have taken $1-0.25^n$
$endgroup$
– Doug M
Mar 29 at 22:33
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Well, each step you have $frac 14$ of what was previously left left. That is after $k$ steps you will have $(frac 14)^k$ left.
You take $frac 34$ of $(frac 14)^k$ and add it to what you had.
So
Step 0: You have $frac 34$ of the pie and $frac 14$ left.
Step 1: You have $frac 34 + frac 34times frac 14$ and you have $(frac 14)^2$ left.
Step 2: You have $frac 34 + frac 34 times frac 14 + frac 34times (frac 14)^2$ and you have $(frac 14)^3$ left .. and so on.
Step $k$: You will have $frac 34 + frac 34 times frac 14 + .... + frac 34 times (frac 14)^k$ and you will $(frac 14)^k+1$ of the pie left.
Thus $sumlimits_k=0^n frac 34times (frac 14)^k=frac 34sumlimits_k=0^n (frac 14)^k$.
.... or if you prefer $3 sumlimits_k=1^n+1 (frac 14)^k$
Now maybe you do, or do not, know how to express terms such a $1 + x + x^2 + x^3 + ........ + x^n$?
answered Mar 29 at 22:41
fleabloodfleablood
73.8k22891
$endgroup$
add a comment |
$begingroup$
You could write this as a recursive sequence:
$$casesa_0 = 0 \
a_n = a_n-1 + 0.75(1 - a_n-1) = 0.75 + 0.25 a_n-1$$
This notation is kind of convenient because you can clearly see that the limit will eventually be the whole pie:
$$L = 0.75 + 0.25 L implies 4 L = 3 + L implies L = 1$$
answered Mar 29 at 22:41
gdepaulgdepaul
1537
$endgroup$
$begingroup$
Adn here comes the problem ! In a life time, we shall not be able to do it an infinite number of times. So, there is still a piece of pie left ! Cheers :-)
$endgroup$
– Claude Leibovici
Mar 30 at 6:03
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Well, each step you have $frac 14$ of what was previously left left. That is after $k$ steps you will have $(frac 14)^k$ left.
You take $frac 34$ of $(frac 14)^k$ and add it to what you had.
So
Step 0: You have $frac 34$ of the pie and $frac 14$ left.
Step 1: You have $frac 34 + frac 34times frac 14$ and you have $(frac 14)^2$ left.
Step 2: You have $frac 34 + frac 34 times frac 14 + frac 34times (frac 14)^2$ and you have $(frac 14)^3$ left .. and so on.
Step $k$: You will have $frac 34 + frac 34 times frac 14 + .... + frac 34 times (frac 14)^k$ and you will $(frac 14)^k+1$ of the pie left.
Thus $sumlimits_k=0^n frac 34times (frac 14)^k=frac 34sumlimits_k=0^n (frac 14)^k$.
.... or if you prefer $3 sumlimits_k=1^n+1 (frac 14)^k$
Now maybe you do, or do not, know how to express terms such a $1 + x + x^2 + x^3 + ........ + x^n$?
answered Mar 29 at 22:41
fleabloodfleablood
73.8k22891
$endgroup$
add a comment |
$begingroup$
Well, each step you have $frac 14$ of what was previously left left. That is after $k$ steps you will have $(frac 14)^k$ left.
You take $frac 34$ of $(frac 14)^k$ and add it to what you had.
So
Step 0: You have $frac 34$ of the pie and $frac 14$ left.
Step 1: You have $frac 34 + frac 34times frac 14$ and you have $(frac 14)^2$ left.
Step 2: You have $frac 34 + frac 34 times frac 14 + frac 34times (frac 14)^2$ and you have $(frac 14)^3$ left .. and so on.
Step $k$: You will have $frac 34 + frac 34 times frac 14 + .... + frac 34 times (frac 14)^k$ and you will $(frac 14)^k+1$ of the pie left.
Thus $sumlimits_k=0^n frac 34times (frac 14)^k=frac 34sumlimits_k=0^n (frac 14)^k$.
.... or if you prefer $3 sumlimits_k=1^n+1 (frac 14)^k$
Now maybe you do, or do not, know how to express terms such a $1 + x + x^2 + x^3 + ........ + x^n$?
answered Mar 29 at 22:41
fleabloodfleablood
73.8k22891
$endgroup$
add a comment |
$begingroup$
Well, each step you have $frac 14$ of what was previously left left. That is after $k$ steps you will have $(frac 14)^k$ left.
You take $frac 34$ of $(frac 14)^k$ and add it to what you had.
So
Step 0: You have $frac 34$ of the pie and $frac 14$ left.
Step 1: You have $frac 34 + frac 34times frac 14$ and you have $(frac 14)^2$ left.
Step 2: You have $frac 34 + frac 34 times frac 14 + frac 34times (frac 14)^2$ and you have $(frac 14)^3$ left .. and so on.
Step $k$: You will have $frac 34 + frac 34 times frac 14 + .... + frac 34 times (frac 14)^k$ and you will $(frac 14)^k+1$ of the pie left.
Thus $sumlimits_k=0^n frac 34times (frac 14)^k=frac 34sumlimits_k=0^n (frac 14)^k$.
.... or if you prefer $3 sumlimits_k=1^n+1 (frac 14)^k$
Now maybe you do, or do not, know how to express terms such a $1 + x + x^2 + x^3 + ........ + x^n$?
answered Mar 29 at 22:41
fleabloodfleablood
73.8k22891
$endgroup$
Well, each step you have $frac 14$ of what was previously left left. That is after $k$ steps you will have $(frac 14)^k$ left.
You take $frac 34$ of $(frac 14)^k$ and add it to what you had.
So
Step 0: You have $frac 34$ of the pie and $frac 14$ left.
Step 1: You have $frac 34 + frac 34times frac 14$ and you have $(frac 14)^2$ left.
Step 2: You have $frac 34 + frac 34 times frac 14 + frac 34times (frac 14)^2$ and you have $(frac 14)^3$ left .. and so on.
Step $k$: You will have $frac 34 + frac 34 times frac 14 + .... + frac 34 times (frac 14)^k$ and you will $(frac 14)^k+1$ of the pie left.
Thus $sumlimits_k=0^n frac 34times (frac 14)^k=frac 34sumlimits_k=0^n (frac 14)^k$.
.... or if you prefer $3 sumlimits_k=1^n+1 (frac 14)^k$
Now maybe you do, or do not, know how to express terms such a $1 + x + x^2 + x^3 + ........ + x^n$?
answered Mar 29 at 22:41
fleabloodfleablood
73.8k22891
answered Mar 29 at 22:41
fleabloodfleablood
73.8k22891
answered Mar 29 at 22:41
fleabloodfleablood
73.8k22891
answered Mar 29 at 22:41
fleabloodfleablood
73.8k22891
73.8k22891
add a comment |
add a comment |
$begingroup$
You could write this as a recursive sequence:
$$casesa_0 = 0 \
a_n = a_n-1 + 0.75(1 - a_n-1) = 0.75 + 0.25 a_n-1$$
This notation is kind of convenient because you can clearly see that the limit will eventually be the whole pie:
$$L = 0.75 + 0.25 L implies 4 L = 3 + L implies L = 1$$
answered Mar 29 at 22:41
gdepaulgdepaul
1537
$endgroup$
$begingroup$
Adn here comes the problem ! In a life time, we shall not be able to do it an infinite number of times. So, there is still a piece of pie left ! Cheers :-)
$endgroup$
– Claude Leibovici
Mar 30 at 6:03
add a comment |
$begingroup$
You could write this as a recursive sequence:
$$casesa_0 = 0 \
a_n = a_n-1 + 0.75(1 - a_n-1) = 0.75 + 0.25 a_n-1$$
This notation is kind of convenient because you can clearly see that the limit will eventually be the whole pie:
$$L = 0.75 + 0.25 L implies 4 L = 3 + L implies L = 1$$
answered Mar 29 at 22:41
gdepaulgdepaul
1537
$endgroup$
$begingroup$
Adn here comes the problem ! In a life time, we shall not be able to do it an infinite number of times. So, there is still a piece of pie left ! Cheers :-)
$endgroup$
– Claude Leibovici
Mar 30 at 6:03
add a comment |
$begingroup$
You could write this as a recursive sequence:
$$casesa_0 = 0 \
a_n = a_n-1 + 0.75(1 - a_n-1) = 0.75 + 0.25 a_n-1$$
This notation is kind of convenient because you can clearly see that the limit will eventually be the whole pie:
$$L = 0.75 + 0.25 L implies 4 L = 3 + L implies L = 1$$
answered Mar 29 at 22:41
gdepaulgdepaul
1537
$endgroup$
You could write this as a recursive sequence:
$$casesa_0 = 0 \
a_n = a_n-1 + 0.75(1 - a_n-1) = 0.75 + 0.25 a_n-1$$
This notation is kind of convenient because you can clearly see that the limit will eventually be the whole pie:
$$L = 0.75 + 0.25 L implies 4 L = 3 + L implies L = 1$$
answered Mar 29 at 22:41
gdepaulgdepaul
1537
answered Mar 29 at 22:41
gdepaulgdepaul
1537
answered Mar 29 at 22:41
gdepaulgdepaul
1537
answered Mar 29 at 22:41
gdepaulgdepaul
1537
1537
$begingroup$
Adn here comes the problem ! In a life time, we shall not be able to do it an infinite number of times. So, there is still a piece of pie left ! Cheers :-)
$endgroup$
– Claude Leibovici
Mar 30 at 6:03
add a comment |
$begingroup$
Adn here comes the problem ! In a life time, we shall not be able to do it an infinite number of times. So, there is still a piece of pie left ! Cheers :-)
$endgroup$
– Claude Leibovici
Mar 30 at 6:03
$begingroup$
Adn here comes the problem ! In a life time, we shall not be able to do it an infinite number of times. So, there is still a piece of pie left ! Cheers :-)
$endgroup$
– Claude Leibovici
Mar 30 at 6:03
$begingroup$
Adn here comes the problem ! In a life time, we shall not be able to do it an infinite number of times. So, there is still a piece of pie left ! Cheers :-)
$endgroup$
– Claude Leibovici
Mar 30 at 6:03
add a comment |
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$begingroup$
Start by writing out the steps. Look for a pattern.
$endgroup$
– Xander Henderson
Mar 29 at 22:33
1
$begingroup$
Think of the pie you don't take. Left behind there is $(0.25)^n$ of the original pie. Which means that you have taken $1-0.25^n$
$endgroup$
– Doug M
Mar 29 at 22:33