Curve with value $0$ at $x=0$, max value $1$ at $x=1$, then decay approximating $frac1x$?

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I want to create a curve with the above properties, and which is odd in that it is continuous, and produces a negative version of itself for negative $x$. A picture may help give the general idea (but please note it's not accurate):

This is created by shifting the normal (probability) curve to centre on $x=1$, then subtracting another normal curve shifted to centre on $x=-1$, using this basic formula for a normal curve:

$$frace^frac-x^22sqrt (2pi)$$

I then stretched the curve vertically by $fracsqrt (2pi)1-frac1e^2$ to give the curve above.

Unfortunately, it's not quite what I am looking for.

1. I want maxima and minima at $x=1$ and $x=-1$


2. I want the value of the function at those points to be $1$ and $-1$


3. I want the decay after that point to follow (at least very roughly) $frac1x$ - i.e., far less rapid

Any suggestions?

algebra-precalculus graphing-functions curves

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edited Jul 17 at 14:33

Parcly Taxel

33.6k136588

asked Jul 17 at 14:25

Richard Burke-Ward

1448

add a comment | 

up vote
2
down vote

favorite

1

I want to create a curve with the above properties, and which is odd in that it is continuous, and produces a negative version of itself for negative $x$. A picture may help give the general idea (but please note it's not accurate):

This is created by shifting the normal (probability) curve to centre on $x=1$, then subtracting another normal curve shifted to centre on $x=-1$, using this basic formula for a normal curve:

$$frace^frac-x^22sqrt (2pi)$$

I then stretched the curve vertically by $fracsqrt (2pi)1-frac1e^2$ to give the curve above.

Unfortunately, it's not quite what I am looking for.

1. I want maxima and minima at $x=1$ and $x=-1$


2. I want the value of the function at those points to be $1$ and $-1$


3. I want the decay after that point to follow (at least very roughly) $frac1x$ - i.e., far less rapid

Any suggestions?

algebra-precalculus graphing-functions curves

share|cite|improve this question

edited Jul 17 at 14:33

Parcly Taxel

33.6k136588

asked Jul 17 at 14:25

Richard Burke-Ward

1448

add a comment | 

up vote
2
down vote

favorite

1

up vote
2
down vote

favorite

1

1

I want to create a curve with the above properties, and which is odd in that it is continuous, and produces a negative version of itself for negative $x$. A picture may help give the general idea (but please note it's not accurate):

This is created by shifting the normal (probability) curve to centre on $x=1$, then subtracting another normal curve shifted to centre on $x=-1$, using this basic formula for a normal curve:

$$frace^frac-x^22sqrt (2pi)$$

I then stretched the curve vertically by $fracsqrt (2pi)1-frac1e^2$ to give the curve above.

Unfortunately, it's not quite what I am looking for.

1. I want maxima and minima at $x=1$ and $x=-1$


2. I want the value of the function at those points to be $1$ and $-1$


3. I want the decay after that point to follow (at least very roughly) $frac1x$ - i.e., far less rapid

Any suggestions?

algebra-precalculus graphing-functions curves

share|cite|improve this question

edited Jul 17 at 14:33

Parcly Taxel

33.6k136588

asked Jul 17 at 14:25

Richard Burke-Ward

1448

I want to create a curve with the above properties, and which is odd in that it is continuous, and produces a negative version of itself for negative $x$. A picture may help give the general idea (but please note it's not accurate):

This is created by shifting the normal (probability) curve to centre on $x=1$, then subtracting another normal curve shifted to centre on $x=-1$, using this basic formula for a normal curve:

$$frace^frac-x^22sqrt (2pi)$$

I then stretched the curve vertically by $fracsqrt (2pi)1-frac1e^2$ to give the curve above.

Unfortunately, it's not quite what I am looking for.

1. I want maxima and minima at $x=1$ and $x=-1$


2. I want the value of the function at those points to be $1$ and $-1$


3. I want the decay after that point to follow (at least very roughly) $frac1x$ - i.e., far less rapid

Any suggestions?

algebra-precalculus graphing-functions curves

share|cite|improve this question

edited Jul 17 at 14:33

Parcly Taxel

33.6k136588

asked Jul 17 at 14:25

Richard Burke-Ward

1448

share|cite|improve this question

share|cite|improve this question

edited Jul 17 at 14:33

Parcly Taxel

33.6k136588

edited Jul 17 at 14:33

Parcly Taxel

33.6k136588

edited Jul 17 at 14:33

Parcly Taxel

33.6k136588

33.6k136588

asked Jul 17 at 14:25

Richard Burke-Ward

1448

asked Jul 17 at 14:25

Richard Burke-Ward

1448

asked Jul 17 at 14:25

Richard Burke-Ward

1448

1448

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
3
down vote



accepted

The simplest function available that satisfies the given properties is a rational function:
$$frac2x1+x^2$$
That it is odd (and so has $f(0)=0$) comes from the $x$ in the numerator. That it decays at the rate of $frac1x$ comes from comparing the degrees of the polynomials above and below (1 vs. 2). That it has a maximum of $f(1)=1$ can be verified by differentiating the function.

share|cite|improve this answer

edited Jul 17 at 14:30

Did

242k23208443

answered Jul 17 at 14:29

Parcly Taxel

33.6k136588

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  +1. Added a graph (since I was in the midst of posting basically the same answer). Delete it if you object.
  – Did
  Jul 17 at 14:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Far simpler! Appreciated, and marked as answered.
  – Richard Burke-Ward
  Jul 17 at 14:35
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

The simplest function available that satisfies the given properties is a rational function:
$$frac2x1+x^2$$
That it is odd (and so has $f(0)=0$) comes from the $x$ in the numerator. That it decays at the rate of $frac1x$ comes from comparing the degrees of the polynomials above and below (1 vs. 2). That it has a maximum of $f(1)=1$ can be verified by differentiating the function.

share|cite|improve this answer

edited Jul 17 at 14:30

Did

242k23208443

answered Jul 17 at 14:29

Parcly Taxel

33.6k136588

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  +1. Added a graph (since I was in the midst of posting basically the same answer). Delete it if you object.
  – Did
  Jul 17 at 14:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Far simpler! Appreciated, and marked as answered.
  – Richard Burke-Ward
  Jul 17 at 14:35
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

The simplest function available that satisfies the given properties is a rational function:
$$frac2x1+x^2$$
That it is odd (and so has $f(0)=0$) comes from the $x$ in the numerator. That it decays at the rate of $frac1x$ comes from comparing the degrees of the polynomials above and below (1 vs. 2). That it has a maximum of $f(1)=1$ can be verified by differentiating the function.

share|cite|improve this answer

edited Jul 17 at 14:30

Did

242k23208443

answered Jul 17 at 14:29

Parcly Taxel

33.6k136588

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  +1. Added a graph (since I was in the midst of posting basically the same answer). Delete it if you object.
  – Did
  Jul 17 at 14:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Far simpler! Appreciated, and marked as answered.
  – Richard Burke-Ward
  Jul 17 at 14:35
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

The simplest function available that satisfies the given properties is a rational function:
$$frac2x1+x^2$$
That it is odd (and so has $f(0)=0$) comes from the $x$ in the numerator. That it decays at the rate of $frac1x$ comes from comparing the degrees of the polynomials above and below (1 vs. 2). That it has a maximum of $f(1)=1$ can be verified by differentiating the function.

share|cite|improve this answer

edited Jul 17 at 14:30

Did

242k23208443

answered Jul 17 at 14:29

Parcly Taxel

33.6k136588

The simplest function available that satisfies the given properties is a rational function:
$$frac2x1+x^2$$
That it is odd (and so has $f(0)=0$) comes from the $x$ in the numerator. That it decays at the rate of $frac1x$ comes from comparing the degrees of the polynomials above and below (1 vs. 2). That it has a maximum of $f(1)=1$ can be verified by differentiating the function.

share|cite|improve this answer

edited Jul 17 at 14:30

Did

242k23208443

answered Jul 17 at 14:29

Parcly Taxel

33.6k136588

share|cite|improve this answer

share|cite|improve this answer

edited Jul 17 at 14:30

Did

242k23208443

edited Jul 17 at 14:30

Did

242k23208443

edited Jul 17 at 14:30

Did

242k23208443

242k23208443

answered Jul 17 at 14:29

Parcly Taxel

33.6k136588

answered Jul 17 at 14:29

Parcly Taxel

33.6k136588

answered Jul 17 at 14:29

Parcly Taxel

33.6k136588

33.6k136588

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  +1. Added a graph (since I was in the midst of posting basically the same answer). Delete it if you object.
  – Did
  Jul 17 at 14:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Far simpler! Appreciated, and marked as answered.
  – Richard Burke-Ward
  Jul 17 at 14:35
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  +1. Added a graph (since I was in the midst of posting basically the same answer). Delete it if you object.
  – Did
  Jul 17 at 14:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Far simpler! Appreciated, and marked as answered.
  – Richard Burke-Ward
  Jul 17 at 14:35
  

  
  
  
  

1

1

+1. Added a graph (since I was in the midst of posting basically the same answer). Delete it if you object.
– Did
Jul 17 at 14:30

+1. Added a graph (since I was in the midst of posting basically the same answer). Delete it if you object.
– Did
Jul 17 at 14:30

Far simpler! Appreciated, and marked as answered.
– Richard Burke-Ward
Jul 17 at 14:35

Far simpler! Appreciated, and marked as answered.
– Richard Burke-Ward
Jul 17 at 14:35

add a comment | 

 

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