Parity of an infinite exponential function (What shape is $y=x^infty$)
$begingroup$
When you have functions of the form $x^c$, the shape of the graph is symmetric with even integers for c (└┘-shape) and for odd integers it is non-symmetric (┌┘-shape)
When dealing with limits as c approaches infinity, how would you plot this function $y=x^c$, as the shape of the function toggles as c goes from even to odd. Is infinity even or odd (obviously neither).
Additionally, the functions do all sorts of wonky things in the complex plain when c is non-integer. So since the plotting of this function $y=x^c$ is so volatile as c changes, how would you plot what the shape of the function looks like as c approaches ∞
calculus limits graphing-functions infinity parity
asked Dec 25 '18 at 20:01
Albert RenshawAlbert Renshaw
7421627
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add a comment |
$begingroup$
When you have functions of the form $x^c$, the shape of the graph is symmetric with even integers for c (└┘-shape) and for odd integers it is non-symmetric (┌┘-shape)
When dealing with limits as c approaches infinity, how would you plot this function $y=x^c$, as the shape of the function toggles as c goes from even to odd. Is infinity even or odd (obviously neither).
Additionally, the functions do all sorts of wonky things in the complex plain when c is non-integer. So since the plotting of this function $y=x^c$ is so volatile as c changes, how would you plot what the shape of the function looks like as c approaches ∞
calculus limits graphing-functions infinity parity
asked Dec 25 '18 at 20:01
Albert RenshawAlbert Renshaw
7421627
$endgroup$
add a comment |
$begingroup$
When you have functions of the form $x^c$, the shape of the graph is symmetric with even integers for c (└┘-shape) and for odd integers it is non-symmetric (┌┘-shape)
When dealing with limits as c approaches infinity, how would you plot this function $y=x^c$, as the shape of the function toggles as c goes from even to odd. Is infinity even or odd (obviously neither).
Additionally, the functions do all sorts of wonky things in the complex plain when c is non-integer. So since the plotting of this function $y=x^c$ is so volatile as c changes, how would you plot what the shape of the function looks like as c approaches ∞
calculus limits graphing-functions infinity parity
asked Dec 25 '18 at 20:01
Albert RenshawAlbert Renshaw
7421627
$endgroup$
When you have functions of the form $x^c$, the shape of the graph is symmetric with even integers for c (└┘-shape) and for odd integers it is non-symmetric (┌┘-shape)
When dealing with limits as c approaches infinity, how would you plot this function $y=x^c$, as the shape of the function toggles as c goes from even to odd. Is infinity even or odd (obviously neither).
Additionally, the functions do all sorts of wonky things in the complex plain when c is non-integer. So since the plotting of this function $y=x^c$ is so volatile as c changes, how would you plot what the shape of the function looks like as c approaches ∞
calculus limits graphing-functions infinity parity
calculus limits graphing-functions infinity parity
asked Dec 25 '18 at 20:01
Albert RenshawAlbert Renshaw
7421627
asked Dec 25 '18 at 20:01
Albert RenshawAlbert Renshaw
7421627
edited Dec 25 '18 at 20:20
Albert Renshaw
asked Dec 25 '18 at 20:01
Albert RenshawAlbert Renshaw
7421627
asked Dec 25 '18 at 20:01
Albert RenshawAlbert Renshaw
7421627
asked Dec 25 '18 at 20:01
Albert RenshawAlbert Renshaw
7421627
7421627
add a comment |
add a comment |
1 Answer
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$begingroup$
You can't really talk about setting $c=infty$, as that is not a number. You can discuss what the limit of the function might look like and then plot that, but you then would need to discuss what the valid outputs of the limit of a sequence of functions could be; in particular, if you allow for the limiting process to result in partial functions and/or functions on the extended real line, we can safely declare that the "answer" would be
$$
y_infty(x) =
begin{cases}
text{undefined} & xleq-1\
0 & |x| < 1 \
1 & x = 1\
infty & x>1
end{cases}
$$
though even this is debatable, as I have implicitly assumed that $c$ is only taking on integral values. If you let the parameter $c$ vary continuously, then a completely different problem arises, and we would be undefined on all negative inputs, rather than those less than or equal to $-1$.
answered Dec 25 '18 at 20:11
ImNotTheGuyImNotTheGuy
38516
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add a comment |
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1 Answer
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active
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1 Answer
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active
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active
oldest
votes
$begingroup$
You can't really talk about setting $c=infty$, as that is not a number. You can discuss what the limit of the function might look like and then plot that, but you then would need to discuss what the valid outputs of the limit of a sequence of functions could be; in particular, if you allow for the limiting process to result in partial functions and/or functions on the extended real line, we can safely declare that the "answer" would be
$$
y_infty(x) =
begin{cases}
text{undefined} & xleq-1\
0 & |x| < 1 \
1 & x = 1\
infty & x>1
end{cases}
$$
though even this is debatable, as I have implicitly assumed that $c$ is only taking on integral values. If you let the parameter $c$ vary continuously, then a completely different problem arises, and we would be undefined on all negative inputs, rather than those less than or equal to $-1$.
answered Dec 25 '18 at 20:11
ImNotTheGuyImNotTheGuy
38516
$endgroup$
add a comment |
$begingroup$
You can't really talk about setting $c=infty$, as that is not a number. You can discuss what the limit of the function might look like and then plot that, but you then would need to discuss what the valid outputs of the limit of a sequence of functions could be; in particular, if you allow for the limiting process to result in partial functions and/or functions on the extended real line, we can safely declare that the "answer" would be
$$
y_infty(x) =
begin{cases}
text{undefined} & xleq-1\
0 & |x| < 1 \
1 & x = 1\
infty & x>1
end{cases}
$$
though even this is debatable, as I have implicitly assumed that $c$ is only taking on integral values. If you let the parameter $c$ vary continuously, then a completely different problem arises, and we would be undefined on all negative inputs, rather than those less than or equal to $-1$.
answered Dec 25 '18 at 20:11
ImNotTheGuyImNotTheGuy
38516
$endgroup$
add a comment |
$begingroup$
You can't really talk about setting $c=infty$, as that is not a number. You can discuss what the limit of the function might look like and then plot that, but you then would need to discuss what the valid outputs of the limit of a sequence of functions could be; in particular, if you allow for the limiting process to result in partial functions and/or functions on the extended real line, we can safely declare that the "answer" would be
$$
y_infty(x) =
begin{cases}
text{undefined} & xleq-1\
0 & |x| < 1 \
1 & x = 1\
infty & x>1
end{cases}
$$
though even this is debatable, as I have implicitly assumed that $c$ is only taking on integral values. If you let the parameter $c$ vary continuously, then a completely different problem arises, and we would be undefined on all negative inputs, rather than those less than or equal to $-1$.
answered Dec 25 '18 at 20:11
ImNotTheGuyImNotTheGuy
38516
$endgroup$
You can't really talk about setting $c=infty$, as that is not a number. You can discuss what the limit of the function might look like and then plot that, but you then would need to discuss what the valid outputs of the limit of a sequence of functions could be; in particular, if you allow for the limiting process to result in partial functions and/or functions on the extended real line, we can safely declare that the "answer" would be
$$
y_infty(x) =
begin{cases}
text{undefined} & xleq-1\
0 & |x| < 1 \
1 & x = 1\
infty & x>1
end{cases}
$$
though even this is debatable, as I have implicitly assumed that $c$ is only taking on integral values. If you let the parameter $c$ vary continuously, then a completely different problem arises, and we would be undefined on all negative inputs, rather than those less than or equal to $-1$.
answered Dec 25 '18 at 20:11
ImNotTheGuyImNotTheGuy
38516
answered Dec 25 '18 at 20:11
ImNotTheGuyImNotTheGuy
38516
answered Dec 25 '18 at 20:11
ImNotTheGuyImNotTheGuy
38516
answered Dec 25 '18 at 20:11
ImNotTheGuyImNotTheGuy
38516
38516
add a comment |
add a comment |
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