1. What is the physical significance an inflexion point might have?
$begingroup$
What is this question asking?
So far I've got An Inflection Point is where a curve changes from Concave upward to Concave downward (or vice versa) The second derivative tells us if the slope increases or decreases.
• When the second derivative is positive, the function is concave upward.
• When the second derivative is negative, the function is concave downward.
Could an example of this be looking at the economy over time?
calculus
$endgroup$
add a comment |
$begingroup$
What is this question asking?
So far I've got An Inflection Point is where a curve changes from Concave upward to Concave downward (or vice versa) The second derivative tells us if the slope increases or decreases.
• When the second derivative is positive, the function is concave upward.
• When the second derivative is negative, the function is concave downward.
Could an example of this be looking at the economy over time?
calculus
$endgroup$
$begingroup$
For a physical interpretation: have you thought about motion? If $x$ is the position, $v = dx/dt$ is the velocity and $a = dv/dt = d^2x/dt^2$ is the acceleration. Imagine a situation where $a$ becomes 0 and changes sign, how does this affect the motion?
$endgroup$
– StackTD
Apr 13 '16 at 16:59
$begingroup$
hint: On a smooth curve, there is always an inflection between a minimum and a maximum (or conversely). Economy doesn't provide the best examples, as the curves are irregular.
$endgroup$
– Yves Daoust
Apr 13 '16 at 16:59
add a comment |
$begingroup$
What is this question asking?
So far I've got An Inflection Point is where a curve changes from Concave upward to Concave downward (or vice versa) The second derivative tells us if the slope increases or decreases.
• When the second derivative is positive, the function is concave upward.
• When the second derivative is negative, the function is concave downward.
Could an example of this be looking at the economy over time?
calculus
$endgroup$
What is this question asking?
So far I've got An Inflection Point is where a curve changes from Concave upward to Concave downward (or vice versa) The second derivative tells us if the slope increases or decreases.
• When the second derivative is positive, the function is concave upward.
• When the second derivative is negative, the function is concave downward.
Could an example of this be looking at the economy over time?
calculus
calculus
asked Apr 13 '16 at 16:49
JessicaJessica
3017
3017
$begingroup$
For a physical interpretation: have you thought about motion? If $x$ is the position, $v = dx/dt$ is the velocity and $a = dv/dt = d^2x/dt^2$ is the acceleration. Imagine a situation where $a$ becomes 0 and changes sign, how does this affect the motion?
$endgroup$
– StackTD
Apr 13 '16 at 16:59
$begingroup$
hint: On a smooth curve, there is always an inflection between a minimum and a maximum (or conversely). Economy doesn't provide the best examples, as the curves are irregular.
$endgroup$
– Yves Daoust
Apr 13 '16 at 16:59
add a comment |
$begingroup$
For a physical interpretation: have you thought about motion? If $x$ is the position, $v = dx/dt$ is the velocity and $a = dv/dt = d^2x/dt^2$ is the acceleration. Imagine a situation where $a$ becomes 0 and changes sign, how does this affect the motion?
$endgroup$
– StackTD
Apr 13 '16 at 16:59
$begingroup$
hint: On a smooth curve, there is always an inflection between a minimum and a maximum (or conversely). Economy doesn't provide the best examples, as the curves are irregular.
$endgroup$
– Yves Daoust
Apr 13 '16 at 16:59
$begingroup$
For a physical interpretation: have you thought about motion? If $x$ is the position, $v = dx/dt$ is the velocity and $a = dv/dt = d^2x/dt^2$ is the acceleration. Imagine a situation where $a$ becomes 0 and changes sign, how does this affect the motion?
$endgroup$
– StackTD
Apr 13 '16 at 16:59
$begingroup$
For a physical interpretation: have you thought about motion? If $x$ is the position, $v = dx/dt$ is the velocity and $a = dv/dt = d^2x/dt^2$ is the acceleration. Imagine a situation where $a$ becomes 0 and changes sign, how does this affect the motion?
$endgroup$
– StackTD
Apr 13 '16 at 16:59
$begingroup$
hint: On a smooth curve, there is always an inflection between a minimum and a maximum (or conversely). Economy doesn't provide the best examples, as the curves are irregular.
$endgroup$
– Yves Daoust
Apr 13 '16 at 16:59
$begingroup$
hint: On a smooth curve, there is always an inflection between a minimum and a maximum (or conversely). Economy doesn't provide the best examples, as the curves are irregular.
$endgroup$
– Yves Daoust
Apr 13 '16 at 16:59
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $c(t)$ represent the distance travelled by a car at time $t$ on a straight road. Suppose $c(t)$ is strictly monotonic on some interval $[t_0-s,t_0+s]$ with $s>0$ and that $c''(t_0)=0.$ Then as $t$ increased from $t_0-s$ to $t_0$, the velocity, which is positive for $tin [t_0-s,t_0),$ decreased to the local minimum velocity $c'(t_0)$, and as $t$ increased from $t_0$ to $t_0+s$ the velocity increased. In particular if $c'(t_0)=0$ then the car slowed to a stop and then accelerated in the same direction. Example :$c(t)=t^3,$ with $t_0=0.$
$endgroup$
add a comment |
$begingroup$
There are many examples of the physical meaning of an inflection point.
If $y$ is the position of an object in one-dimensional motion, the inflection point is where the force on the object changes direction.
Here are some economic applications. If $y$ is the total cost of producing $x$ number of units of a product, the inflection point is where the marginal cost reaches a minimum (or perhaps maximum).
The growth of a company often follows a logistic curve. If $y$ measures the size of a company in any sense, the inflection point is where the growth is at a maximum. Similarly, the inflection point shows the maximum spread of a sickness, which also usually follows a logistic curve.
Here is a real-life example I show in class. I cut the following graph at $x=22$ and ask the class what is happening to the employment. The inflection point of the "current unemployment recession" is at about $x=12$, so that is where the rate switches from getting worse and picking up speed to getting worse but we're about to turn the corner. That point was into Obama's presidency, so after that point Obama was claiming that he was helping the economy even though employment was still getting worse. The politics is debatable but that does follow the graph.
I summarize with a quote from Hugo Rossi. "In the fall of 1972 President Nixon announced that the rate of increase of inflation was decreasing. This was the first time a sitting president used the third derivative to advance his case for reelection."
$endgroup$
add a comment |
$begingroup$
There is an intuitive interpretation of the discrete second derivative as a discrete curvature, or more precisely the positive or negative bending degree of the linear interpolating curve, as shown on the figure (adapted from my lectures). On this figure one finds :
- 1) (upper part: thanks to Taylor expansions) why $f''(a)$ can be interpolated by the following formula (well known in many applied areas):
$$f''(a)equivdfrac{1}{h^2}(f_{k-1}-2f_k+f_{k+1})$$
- 2) (lower part) the geometric interpretation. It is a comparison between the actual value $f_k=f(a)$ and the "mid-value" $m=dfrac{f_{k-1}+f_{k+1}}{2}$ obtained by taking the arithmetical mean between the neighboring values.
Said succinctly, if $f(a)>m$ (resp. $f(a)<m$) the curve bumps upwards (resp. downwards); note that the "scale-factor" $dfrac{1}{h^2}$ has been replaced by $dfrac{1}{2}$, an unimportant transformation as long as we are only interested by the sign of the difference and not its magnitude.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
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$begingroup$
Let $c(t)$ represent the distance travelled by a car at time $t$ on a straight road. Suppose $c(t)$ is strictly monotonic on some interval $[t_0-s,t_0+s]$ with $s>0$ and that $c''(t_0)=0.$ Then as $t$ increased from $t_0-s$ to $t_0$, the velocity, which is positive for $tin [t_0-s,t_0),$ decreased to the local minimum velocity $c'(t_0)$, and as $t$ increased from $t_0$ to $t_0+s$ the velocity increased. In particular if $c'(t_0)=0$ then the car slowed to a stop and then accelerated in the same direction. Example :$c(t)=t^3,$ with $t_0=0.$
$endgroup$
add a comment |
$begingroup$
Let $c(t)$ represent the distance travelled by a car at time $t$ on a straight road. Suppose $c(t)$ is strictly monotonic on some interval $[t_0-s,t_0+s]$ with $s>0$ and that $c''(t_0)=0.$ Then as $t$ increased from $t_0-s$ to $t_0$, the velocity, which is positive for $tin [t_0-s,t_0),$ decreased to the local minimum velocity $c'(t_0)$, and as $t$ increased from $t_0$ to $t_0+s$ the velocity increased. In particular if $c'(t_0)=0$ then the car slowed to a stop and then accelerated in the same direction. Example :$c(t)=t^3,$ with $t_0=0.$
$endgroup$
add a comment |
$begingroup$
Let $c(t)$ represent the distance travelled by a car at time $t$ on a straight road. Suppose $c(t)$ is strictly monotonic on some interval $[t_0-s,t_0+s]$ with $s>0$ and that $c''(t_0)=0.$ Then as $t$ increased from $t_0-s$ to $t_0$, the velocity, which is positive for $tin [t_0-s,t_0),$ decreased to the local minimum velocity $c'(t_0)$, and as $t$ increased from $t_0$ to $t_0+s$ the velocity increased. In particular if $c'(t_0)=0$ then the car slowed to a stop and then accelerated in the same direction. Example :$c(t)=t^3,$ with $t_0=0.$
$endgroup$
Let $c(t)$ represent the distance travelled by a car at time $t$ on a straight road. Suppose $c(t)$ is strictly monotonic on some interval $[t_0-s,t_0+s]$ with $s>0$ and that $c''(t_0)=0.$ Then as $t$ increased from $t_0-s$ to $t_0$, the velocity, which is positive for $tin [t_0-s,t_0),$ decreased to the local minimum velocity $c'(t_0)$, and as $t$ increased from $t_0$ to $t_0+s$ the velocity increased. In particular if $c'(t_0)=0$ then the car slowed to a stop and then accelerated in the same direction. Example :$c(t)=t^3,$ with $t_0=0.$
answered Apr 13 '16 at 17:48
DanielWainfleetDanielWainfleet
34.6k31648
34.6k31648
add a comment |
add a comment |
$begingroup$
There are many examples of the physical meaning of an inflection point.
If $y$ is the position of an object in one-dimensional motion, the inflection point is where the force on the object changes direction.
Here are some economic applications. If $y$ is the total cost of producing $x$ number of units of a product, the inflection point is where the marginal cost reaches a minimum (or perhaps maximum).
The growth of a company often follows a logistic curve. If $y$ measures the size of a company in any sense, the inflection point is where the growth is at a maximum. Similarly, the inflection point shows the maximum spread of a sickness, which also usually follows a logistic curve.
Here is a real-life example I show in class. I cut the following graph at $x=22$ and ask the class what is happening to the employment. The inflection point of the "current unemployment recession" is at about $x=12$, so that is where the rate switches from getting worse and picking up speed to getting worse but we're about to turn the corner. That point was into Obama's presidency, so after that point Obama was claiming that he was helping the economy even though employment was still getting worse. The politics is debatable but that does follow the graph.
I summarize with a quote from Hugo Rossi. "In the fall of 1972 President Nixon announced that the rate of increase of inflation was decreasing. This was the first time a sitting president used the third derivative to advance his case for reelection."
$endgroup$
add a comment |
$begingroup$
There are many examples of the physical meaning of an inflection point.
If $y$ is the position of an object in one-dimensional motion, the inflection point is where the force on the object changes direction.
Here are some economic applications. If $y$ is the total cost of producing $x$ number of units of a product, the inflection point is where the marginal cost reaches a minimum (or perhaps maximum).
The growth of a company often follows a logistic curve. If $y$ measures the size of a company in any sense, the inflection point is where the growth is at a maximum. Similarly, the inflection point shows the maximum spread of a sickness, which also usually follows a logistic curve.
Here is a real-life example I show in class. I cut the following graph at $x=22$ and ask the class what is happening to the employment. The inflection point of the "current unemployment recession" is at about $x=12$, so that is where the rate switches from getting worse and picking up speed to getting worse but we're about to turn the corner. That point was into Obama's presidency, so after that point Obama was claiming that he was helping the economy even though employment was still getting worse. The politics is debatable but that does follow the graph.
I summarize with a quote from Hugo Rossi. "In the fall of 1972 President Nixon announced that the rate of increase of inflation was decreasing. This was the first time a sitting president used the third derivative to advance his case for reelection."
$endgroup$
add a comment |
$begingroup$
There are many examples of the physical meaning of an inflection point.
If $y$ is the position of an object in one-dimensional motion, the inflection point is where the force on the object changes direction.
Here are some economic applications. If $y$ is the total cost of producing $x$ number of units of a product, the inflection point is where the marginal cost reaches a minimum (or perhaps maximum).
The growth of a company often follows a logistic curve. If $y$ measures the size of a company in any sense, the inflection point is where the growth is at a maximum. Similarly, the inflection point shows the maximum spread of a sickness, which also usually follows a logistic curve.
Here is a real-life example I show in class. I cut the following graph at $x=22$ and ask the class what is happening to the employment. The inflection point of the "current unemployment recession" is at about $x=12$, so that is where the rate switches from getting worse and picking up speed to getting worse but we're about to turn the corner. That point was into Obama's presidency, so after that point Obama was claiming that he was helping the economy even though employment was still getting worse. The politics is debatable but that does follow the graph.
I summarize with a quote from Hugo Rossi. "In the fall of 1972 President Nixon announced that the rate of increase of inflation was decreasing. This was the first time a sitting president used the third derivative to advance his case for reelection."
$endgroup$
There are many examples of the physical meaning of an inflection point.
If $y$ is the position of an object in one-dimensional motion, the inflection point is where the force on the object changes direction.
Here are some economic applications. If $y$ is the total cost of producing $x$ number of units of a product, the inflection point is where the marginal cost reaches a minimum (or perhaps maximum).
The growth of a company often follows a logistic curve. If $y$ measures the size of a company in any sense, the inflection point is where the growth is at a maximum. Similarly, the inflection point shows the maximum spread of a sickness, which also usually follows a logistic curve.
Here is a real-life example I show in class. I cut the following graph at $x=22$ and ask the class what is happening to the employment. The inflection point of the "current unemployment recession" is at about $x=12$, so that is where the rate switches from getting worse and picking up speed to getting worse but we're about to turn the corner. That point was into Obama's presidency, so after that point Obama was claiming that he was helping the economy even though employment was still getting worse. The politics is debatable but that does follow the graph.
I summarize with a quote from Hugo Rossi. "In the fall of 1972 President Nixon announced that the rate of increase of inflation was decreasing. This was the first time a sitting president used the third derivative to advance his case for reelection."
edited Apr 13 '16 at 18:18
answered Apr 13 '16 at 18:04
Rory DaultonRory Daulton
29.4k53254
29.4k53254
add a comment |
add a comment |
$begingroup$
There is an intuitive interpretation of the discrete second derivative as a discrete curvature, or more precisely the positive or negative bending degree of the linear interpolating curve, as shown on the figure (adapted from my lectures). On this figure one finds :
- 1) (upper part: thanks to Taylor expansions) why $f''(a)$ can be interpolated by the following formula (well known in many applied areas):
$$f''(a)equivdfrac{1}{h^2}(f_{k-1}-2f_k+f_{k+1})$$
- 2) (lower part) the geometric interpretation. It is a comparison between the actual value $f_k=f(a)$ and the "mid-value" $m=dfrac{f_{k-1}+f_{k+1}}{2}$ obtained by taking the arithmetical mean between the neighboring values.
Said succinctly, if $f(a)>m$ (resp. $f(a)<m$) the curve bumps upwards (resp. downwards); note that the "scale-factor" $dfrac{1}{h^2}$ has been replaced by $dfrac{1}{2}$, an unimportant transformation as long as we are only interested by the sign of the difference and not its magnitude.
$endgroup$
add a comment |
$begingroup$
There is an intuitive interpretation of the discrete second derivative as a discrete curvature, or more precisely the positive or negative bending degree of the linear interpolating curve, as shown on the figure (adapted from my lectures). On this figure one finds :
- 1) (upper part: thanks to Taylor expansions) why $f''(a)$ can be interpolated by the following formula (well known in many applied areas):
$$f''(a)equivdfrac{1}{h^2}(f_{k-1}-2f_k+f_{k+1})$$
- 2) (lower part) the geometric interpretation. It is a comparison between the actual value $f_k=f(a)$ and the "mid-value" $m=dfrac{f_{k-1}+f_{k+1}}{2}$ obtained by taking the arithmetical mean between the neighboring values.
Said succinctly, if $f(a)>m$ (resp. $f(a)<m$) the curve bumps upwards (resp. downwards); note that the "scale-factor" $dfrac{1}{h^2}$ has been replaced by $dfrac{1}{2}$, an unimportant transformation as long as we are only interested by the sign of the difference and not its magnitude.
$endgroup$
add a comment |
$begingroup$
There is an intuitive interpretation of the discrete second derivative as a discrete curvature, or more precisely the positive or negative bending degree of the linear interpolating curve, as shown on the figure (adapted from my lectures). On this figure one finds :
- 1) (upper part: thanks to Taylor expansions) why $f''(a)$ can be interpolated by the following formula (well known in many applied areas):
$$f''(a)equivdfrac{1}{h^2}(f_{k-1}-2f_k+f_{k+1})$$
- 2) (lower part) the geometric interpretation. It is a comparison between the actual value $f_k=f(a)$ and the "mid-value" $m=dfrac{f_{k-1}+f_{k+1}}{2}$ obtained by taking the arithmetical mean between the neighboring values.
Said succinctly, if $f(a)>m$ (resp. $f(a)<m$) the curve bumps upwards (resp. downwards); note that the "scale-factor" $dfrac{1}{h^2}$ has been replaced by $dfrac{1}{2}$, an unimportant transformation as long as we are only interested by the sign of the difference and not its magnitude.
$endgroup$
There is an intuitive interpretation of the discrete second derivative as a discrete curvature, or more precisely the positive or negative bending degree of the linear interpolating curve, as shown on the figure (adapted from my lectures). On this figure one finds :
- 1) (upper part: thanks to Taylor expansions) why $f''(a)$ can be interpolated by the following formula (well known in many applied areas):
$$f''(a)equivdfrac{1}{h^2}(f_{k-1}-2f_k+f_{k+1})$$
- 2) (lower part) the geometric interpretation. It is a comparison between the actual value $f_k=f(a)$ and the "mid-value" $m=dfrac{f_{k-1}+f_{k+1}}{2}$ obtained by taking the arithmetical mean between the neighboring values.
Said succinctly, if $f(a)>m$ (resp. $f(a)<m$) the curve bumps upwards (resp. downwards); note that the "scale-factor" $dfrac{1}{h^2}$ has been replaced by $dfrac{1}{2}$, an unimportant transformation as long as we are only interested by the sign of the difference and not its magnitude.
answered Apr 13 '16 at 18:51
Jean MarieJean Marie
28.9k41949
28.9k41949
add a comment |
add a comment |
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$begingroup$
For a physical interpretation: have you thought about motion? If $x$ is the position, $v = dx/dt$ is the velocity and $a = dv/dt = d^2x/dt^2$ is the acceleration. Imagine a situation where $a$ becomes 0 and changes sign, how does this affect the motion?
$endgroup$
– StackTD
Apr 13 '16 at 16:59
$begingroup$
hint: On a smooth curve, there is always an inflection between a minimum and a maximum (or conversely). Economy doesn't provide the best examples, as the curves are irregular.
$endgroup$
– Yves Daoust
Apr 13 '16 at 16:59