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How do I notate this?

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0

$begingroup$

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

share|cite|improve this question

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 | 

0

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

edited Mar 29 at 22:34

Xander Henderson

15.1k103556

asked Mar 29 at 22:28

T. OuT. Ou

6

$endgroup$

share|cite|improve this question

edited Mar 29 at 22:34

Xander Henderson

15.1k103556

asked Mar 29 at 22:28

T. OuT. Ou

6

share|cite|improve this question

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

asked Mar 29 at 22:28

T. OuT. Ou

6

asked Mar 29 at 22:28

T. OuT. Ou

6

2 Answers
2

active

oldest

votes

1

$begingroup$

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 $1 + x + x^2 + x^3 + ........ + x^n$?

share|cite|improve this answer

answered Mar 29 at 22:41

fleabloodfleablood

73.8k22891

$endgroup$

add a comment | 

0

$begingroup$

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:

$$L = 0.75 + 0.25 L implies 4 L = 3 + L implies L = 1$$

share|cite|improve this answer

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 | 

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1

$begingroup$

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 $1 + x + x^2 + x^3 + ........ + x^n$?

share|cite|improve this answer

answered Mar 29 at 22:41

fleabloodfleablood

73.8k22891

$endgroup$

add a comment | 

1

$begingroup$

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 $1 + x + x^2 + x^3 + ........ + x^n$?

share|cite|improve this answer

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 $1 + x + x^2 + x^3 + ........ + x^n$?

share|cite|improve this answer

answered Mar 29 at 22:41

fleabloodfleablood

73.8k22891

$endgroup$

share|cite|improve this answer

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

0

$begingroup$

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:

$$L = 0.75 + 0.25 L implies 4 L = 3 + L implies L = 1$$

share|cite|improve this answer

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 | 

0

$begingroup$

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:

$$L = 0.75 + 0.25 L implies 4 L = 3 + L implies L = 1$$

share|cite|improve this answer

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:

$$L = 0.75 + 0.25 L implies 4 L = 3 + L implies L = 1$$

share|cite|improve this answer

answered Mar 29 at 22:41

gdepaulgdepaul

1537

$endgroup$

share|cite|improve this answer

answered Mar 29 at 22:41

gdepaulgdepaul

1537

answered Mar 29 at 22:41

gdepaulgdepaul

1537

answered Mar 29 at 22:41

gdepaulgdepaul

1537