What is the formula for exponential growth with a decay rate?

Clash Royale CLAN TAG#URR8PPP

up vote
0
down vote

favorite

Exponential growth can be modeled as

$$
b (1+r)^N
$$

For $b$ your starting quantity, $(1+r)$ your rate of growth, and $N$ the number of periods. But for $N to infty$, this formula can get out of control.

Is there a traditional way of controlling for this by factoring in some notion of a decay factor (so that for periods $N$ past some threshold, you stop growing asymptotically)?

exponential-function

share|cite|improve this question

asked Jul 20 at 12:17

user1770201

1,29221134

add a comment | 

up vote
0
down vote

favorite

Exponential growth can be modeled as

$$
b (1+r)^N
$$

For $b$ your starting quantity, $(1+r)$ your rate of growth, and $N$ the number of periods. But for $N to infty$, this formula can get out of control.

Is there a traditional way of controlling for this by factoring in some notion of a decay factor (so that for periods $N$ past some threshold, you stop growing asymptotically)?

exponential-function

share|cite|improve this question

asked Jul 20 at 12:17

user1770201

1,29221134

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Exponential growth can be modeled as

$$
b (1+r)^N
$$

For $b$ your starting quantity, $(1+r)$ your rate of growth, and $N$ the number of periods. But for $N to infty$, this formula can get out of control.

Is there a traditional way of controlling for this by factoring in some notion of a decay factor (so that for periods $N$ past some threshold, you stop growing asymptotically)?

exponential-function

share|cite|improve this question

asked Jul 20 at 12:17

user1770201

1,29221134

Exponential growth can be modeled as

$$
b (1+r)^N
$$

For $b$ your starting quantity, $(1+r)$ your rate of growth, and $N$ the number of periods. But for $N to infty$, this formula can get out of control.

Is there a traditional way of controlling for this by factoring in some notion of a decay factor (so that for periods $N$ past some threshold, you stop growing asymptotically)?

exponential-function

share|cite|improve this question

asked Jul 20 at 12:17

user1770201

1,29221134

share|cite|improve this question

share|cite|improve this question

asked Jul 20 at 12:17

user1770201

1,29221134

asked Jul 20 at 12:17

user1770201

1,29221134

asked Jul 20 at 12:17

user1770201

1,29221134

1,29221134

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
0
down vote

The simplest extension of the exponential equation

$fracdxdt = rx$

is the logistic equation

$fracdxdt = rx(1-fracxC)$

where the rate of growth decreases as $x$ approaches $C$.

This differential equation has solution

$x(t)=fracCx(0)e^rtC+x(0)(e^rt-1)$

Interestingly, the discrete time step equivalent of the logistic equation (known as the logistic map) can exhibit chaotic behaviour.

share|cite|improve this answer

answered Jul 20 at 15:16

gandalf61

5,684522

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2857579%2fwhat-is-the-formula-for-exponential-growth-with-a-decay-rate%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

The simplest extension of the exponential equation

$fracdxdt = rx$

is the logistic equation

$fracdxdt = rx(1-fracxC)$

where the rate of growth decreases as $x$ approaches $C$.

This differential equation has solution

$x(t)=fracCx(0)e^rtC+x(0)(e^rt-1)$

Interestingly, the discrete time step equivalent of the logistic equation (known as the logistic map) can exhibit chaotic behaviour.

share|cite|improve this answer

answered Jul 20 at 15:16

gandalf61

5,684522

add a comment | 

up vote
0
down vote

The simplest extension of the exponential equation

$fracdxdt = rx$

is the logistic equation

$fracdxdt = rx(1-fracxC)$

where the rate of growth decreases as $x$ approaches $C$.

This differential equation has solution

$x(t)=fracCx(0)e^rtC+x(0)(e^rt-1)$

Interestingly, the discrete time step equivalent of the logistic equation (known as the logistic map) can exhibit chaotic behaviour.

share|cite|improve this answer

answered Jul 20 at 15:16

gandalf61

5,684522

add a comment | 

up vote
0
down vote

up vote
0
down vote

The simplest extension of the exponential equation

$fracdxdt = rx$

is the logistic equation

$fracdxdt = rx(1-fracxC)$

where the rate of growth decreases as $x$ approaches $C$.

This differential equation has solution

$x(t)=fracCx(0)e^rtC+x(0)(e^rt-1)$

Interestingly, the discrete time step equivalent of the logistic equation (known as the logistic map) can exhibit chaotic behaviour.

share|cite|improve this answer

answered Jul 20 at 15:16

gandalf61

5,684522

The simplest extension of the exponential equation

$fracdxdt = rx$

is the logistic equation

$fracdxdt = rx(1-fracxC)$

where the rate of growth decreases as $x$ approaches $C$.

This differential equation has solution

$x(t)=fracCx(0)e^rtC+x(0)(e^rt-1)$

Interestingly, the discrete time step equivalent of the logistic equation (known as the logistic map) can exhibit chaotic behaviour.

share|cite|improve this answer

answered Jul 20 at 15:16

gandalf61

5,684522

share|cite|improve this answer

share|cite|improve this answer

answered Jul 20 at 15:16

gandalf61

5,684522

answered Jul 20 at 15:16

gandalf61

5,684522

answered Jul 20 at 15:16

gandalf61

5,684522

5,684522

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2857579%2fwhat-is-the-formula-for-exponential-growth-with-a-decay-rate%23new-answer', 'question_page');

);

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email