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Can you prove the power rule for irrational exponents without invoking $e$?
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The power rule states that for any real number $r$,
$$fracddxx^r=rx^r-1$$
Now one common way to prove this is to use the definition $x^r=e^rln x$, where $e^x$ is defined as the inverse function of $ln x$, which is in turn defined as $int_1^xfracdtt$.
But this puts the cart before the horse, because students typically learn differential calculus before integral calculus. And there is a perfectly good definition of exponentiation of real numbers that does not rely on integral calculus:
$$x^r=lim_qrightarrow r x^q$$
where $q$ is a variable that ranges over the rational numbers.
So my question is, if we use this definition, and we take it for granted that $fracddxx^q=qx^q-1$ holds true for rational numbers (which can be easily proven without invoking $e$), then can we prove the power rule for real exponents without invoking $e$?
EDIT: Here’s a more precise formulation of the definition above. If $r$ is a real number, we say that $x^r = L$ if for any $epsilon>0$ there exists a $delta>0$ such that for any rational number $q$ such that $|q-x| < delta$, we have $|x^q-L|<epsilon$.
calculus real-analysis limits derivatives exponentiation
asked Jul 31 at 15:20
Keshav Srinivasan
1,74511338
 |Â
show 11 more comments
up vote
6
down vote
favorite
The power rule states that for any real number $r$,
$$fracddxx^r=rx^r-1$$
Now one common way to prove this is to use the definition $x^r=e^rln x$, where $e^x$ is defined as the inverse function of $ln x$, which is in turn defined as $int_1^xfracdtt$.
But this puts the cart before the horse, because students typically learn differential calculus before integral calculus. And there is a perfectly good definition of exponentiation of real numbers that does not rely on integral calculus:
$$x^r=lim_qrightarrow r x^q$$
where $q$ is a variable that ranges over the rational numbers.
So my question is, if we use this definition, and we take it for granted that $fracddxx^q=qx^q-1$ holds true for rational numbers (which can be easily proven without invoking $e$), then can we prove the power rule for real exponents without invoking $e$?
EDIT: Here’s a more precise formulation of the definition above. If $r$ is a real number, we say that $x^r = L$ if for any $epsilon>0$ there exists a $delta>0$ such that for any rational number $q$ such that $|q-x| < delta$, we have $|x^q-L|<epsilon$.
calculus real-analysis limits derivatives exponentiation
asked Jul 31 at 15:20
Keshav Srinivasan
1,74511338
Binomial theorem.
– copper.hat
Jul 31 at 15:29
1
@copper.hat Binomial theorem only allows you to derive this for $rin BbbN$
– Michal Dvořák
Jul 31 at 15:30
1
@MichalDvořák That’s not a duplicate at all, that’s about proving properties of the natural logarithm function, my question is about deriving the power rule without talking about $e$ and $ln(x)$ at all.
– Keshav Srinivasan
Jul 31 at 15:31
1
@MichalDvořák: The binomial theorem works for more than natural powers. It avoids using integrals which is what the OP was looking for.
– copper.hat
Jul 31 at 15:32
1
@ParamanandSingh Yes, you can. I gave the definition in my question: $x^r=lim_qrightarrow r x^q$ where $q$ is a variable that ranges over the rational numbers. I want to know how to use that definition to prove the power rule.
– Keshav Srinivasan
Jul 31 at 15:35
 |Â
show 11 more comments
up vote
6
down vote
favorite
up vote
6
down vote
favorite
The power rule states that for any real number $r$,
$$fracddxx^r=rx^r-1$$
Now one common way to prove this is to use the definition $x^r=e^rln x$, where $e^x$ is defined as the inverse function of $ln x$, which is in turn defined as $int_1^xfracdtt$.
But this puts the cart before the horse, because students typically learn differential calculus before integral calculus. And there is a perfectly good definition of exponentiation of real numbers that does not rely on integral calculus:
$$x^r=lim_qrightarrow r x^q$$
where $q$ is a variable that ranges over the rational numbers.
So my question is, if we use this definition, and we take it for granted that $fracddxx^q=qx^q-1$ holds true for rational numbers (which can be easily proven without invoking $e$), then can we prove the power rule for real exponents without invoking $e$?
EDIT: Here’s a more precise formulation of the definition above. If $r$ is a real number, we say that $x^r = L$ if for any $epsilon>0$ there exists a $delta>0$ such that for any rational number $q$ such that $|q-x| < delta$, we have $|x^q-L|<epsilon$.
calculus real-analysis limits derivatives exponentiation
asked Jul 31 at 15:20
Keshav Srinivasan
1,74511338
The power rule states that for any real number $r$,
$$fracddxx^r=rx^r-1$$
Now one common way to prove this is to use the definition $x^r=e^rln x$, where $e^x$ is defined as the inverse function of $ln x$, which is in turn defined as $int_1^xfracdtt$.
But this puts the cart before the horse, because students typically learn differential calculus before integral calculus. And there is a perfectly good definition of exponentiation of real numbers that does not rely on integral calculus:
$$x^r=lim_qrightarrow r x^q$$
where $q$ is a variable that ranges over the rational numbers.
So my question is, if we use this definition, and we take it for granted that $fracddxx^q=qx^q-1$ holds true for rational numbers (which can be easily proven without invoking $e$), then can we prove the power rule for real exponents without invoking $e$?
EDIT: Here’s a more precise formulation of the definition above. If $r$ is a real number, we say that $x^r = L$ if for any $epsilon>0$ there exists a $delta>0$ such that for any rational number $q$ such that $|q-x| < delta$, we have $|x^q-L|<epsilon$.
calculus real-analysis limits derivatives exponentiation
asked Jul 31 at 15:20
Keshav Srinivasan
1,74511338
edited Jul 31 at 15:51
asked Jul 31 at 15:20
Keshav Srinivasan
1,74511338
asked Jul 31 at 15:20
Keshav Srinivasan
1,74511338
asked Jul 31 at 15:20
Keshav Srinivasan
1,74511338
1,74511338
Binomial theorem.
– copper.hat
Jul 31 at 15:29
1
@copper.hat Binomial theorem only allows you to derive this for $rin BbbN$
– Michal Dvořák
Jul 31 at 15:30
1
@MichalDvořák That’s not a duplicate at all, that’s about proving properties of the natural logarithm function, my question is about deriving the power rule without talking about $e$ and $ln(x)$ at all.
– Keshav Srinivasan
Jul 31 at 15:31
1
@MichalDvořák: The binomial theorem works for more than natural powers. It avoids using integrals which is what the OP was looking for.
– copper.hat
Jul 31 at 15:32
1
@ParamanandSingh Yes, you can. I gave the definition in my question: $x^r=lim_qrightarrow r x^q$ where $q$ is a variable that ranges over the rational numbers. I want to know how to use that definition to prove the power rule.
– Keshav Srinivasan
Jul 31 at 15:35
 |Â
show 11 more comments
Binomial theorem.
– copper.hat
Jul 31 at 15:29
1
@copper.hat Binomial theorem only allows you to derive this for $rin BbbN$
– Michal Dvořák
Jul 31 at 15:30
1
@MichalDvořák That’s not a duplicate at all, that’s about proving properties of the natural logarithm function, my question is about deriving the power rule without talking about $e$ and $ln(x)$ at all.
– Keshav Srinivasan
Jul 31 at 15:31
1
@MichalDvořák: The binomial theorem works for more than natural powers. It avoids using integrals which is what the OP was looking for.
– copper.hat
Jul 31 at 15:32
1
@ParamanandSingh Yes, you can. I gave the definition in my question: $x^r=lim_qrightarrow r x^q$ where $q$ is a variable that ranges over the rational numbers. I want to know how to use that definition to prove the power rule.
– Keshav Srinivasan
Jul 31 at 15:35
Binomial theorem.
– copper.hat
Jul 31 at 15:29
Binomial theorem.
– copper.hat
Jul 31 at 15:29
1
1
@copper.hat Binomial theorem only allows you to derive this for $rin BbbN$
– Michal Dvořák
Jul 31 at 15:30
@copper.hat Binomial theorem only allows you to derive this for $rin BbbN$
– Michal Dvořák
Jul 31 at 15:30
1
1
@MichalDvořák That’s not a duplicate at all, that’s about proving properties of the natural logarithm function, my question is about deriving the power rule without talking about $e$ and $ln(x)$ at all.
– Keshav Srinivasan
Jul 31 at 15:31
@MichalDvořák That’s not a duplicate at all, that’s about proving properties of the natural logarithm function, my question is about deriving the power rule without talking about $e$ and $ln(x)$ at all.
– Keshav Srinivasan
Jul 31 at 15:31
1
1
@MichalDvořák: The binomial theorem works for more than natural powers. It avoids using integrals which is what the OP was looking for.
– copper.hat
Jul 31 at 15:32
@MichalDvořák: The binomial theorem works for more than natural powers. It avoids using integrals which is what the OP was looking for.
– copper.hat
Jul 31 at 15:32
1
1
@ParamanandSingh Yes, you can. I gave the definition in my question: $x^r=lim_qrightarrow r x^q$ where $q$ is a variable that ranges over the rational numbers. I want to know how to use that definition to prove the power rule.
– Keshav Srinivasan
Jul 31 at 15:35
@ParamanandSingh Yes, you can. I gave the definition in my question: $x^r=lim_qrightarrow r x^q$ where $q$ is a variable that ranges over the rational numbers. I want to know how to use that definition to prove the power rule.
– Keshav Srinivasan
Jul 31 at 15:35
 |Â
show 11 more comments
2 Answers
2
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2
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Yes. In general, if you have a sequence of $C^1$ functions $f_n$ on some interval and functions $f$ and $g$ such that $f_nto f$ pointwise and $f_n'to g$ uniformly, then $f'=g$. (Quick sketch of how to prove this without using integration: fix $x$ and $h$ and pick $n$ so that $f_n$ is sufficiently close to $f$ at $x$ and $x+h$ and $f_n'$ is sufficiently close to $g$ uniformly. Then $fracf(x+h)-f(x)h$ will be close to $fracf_n(x+h)-f_n(x)h$. By the mean value theorem the latter is equal to $f_n'(y)$ for some $y$ between $x+h$ and $x$, and $f_n'(y)$ is close to $g(y)$. Finally, if $h$ is sufficiently small, $g(y)$ will be close to $g(x)$ since $g$ is continuous.)
So, given $rinmathbbR$, pick a sequence of rational numbers $q_n$ converging to $r$ and let $f_n(x)=x^q_n$. Then $f_n(x)$ converges to $f(x)=x^r$ and $f_n'(x)=q_nx^q_n-1$ converges to $g(x)=rx^r-1$. Moreover, it is not hard to check that these convergences are uniform on any compact subset of $(0,infty)$. It thus follows that $f'=g$ on $(0,infty)$.
answered Jul 31 at 16:00
Eric Wofsey
161k12188297
add a comment |Â
up vote
0
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Hint:
$$x^r:=lim_qto r,\qinmathbb Qx^q.$$
Then
$$(x^r)'=lim_hto0fraclim_qto r,\qinmathbb Q((x+h)^q-x^q)h=lim_qto r,\qinmathbb Qlim_hto0frac((x+h)^q-x^q)h=lim_qto r,\qinmathbb Qqx^q-1=rx^r-1.$$
The hard part is to justify the swap of the limits.
answered Jul 31 at 15:58
Yves Daoust
110k665203
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Yes. In general, if you have a sequence of $C^1$ functions $f_n$ on some interval and functions $f$ and $g$ such that $f_nto f$ pointwise and $f_n'to g$ uniformly, then $f'=g$. (Quick sketch of how to prove this without using integration: fix $x$ and $h$ and pick $n$ so that $f_n$ is sufficiently close to $f$ at $x$ and $x+h$ and $f_n'$ is sufficiently close to $g$ uniformly. Then $fracf(x+h)-f(x)h$ will be close to $fracf_n(x+h)-f_n(x)h$. By the mean value theorem the latter is equal to $f_n'(y)$ for some $y$ between $x+h$ and $x$, and $f_n'(y)$ is close to $g(y)$. Finally, if $h$ is sufficiently small, $g(y)$ will be close to $g(x)$ since $g$ is continuous.)
So, given $rinmathbbR$, pick a sequence of rational numbers $q_n$ converging to $r$ and let $f_n(x)=x^q_n$. Then $f_n(x)$ converges to $f(x)=x^r$ and $f_n'(x)=q_nx^q_n-1$ converges to $g(x)=rx^r-1$. Moreover, it is not hard to check that these convergences are uniform on any compact subset of $(0,infty)$. It thus follows that $f'=g$ on $(0,infty)$.
answered Jul 31 at 16:00
Eric Wofsey
161k12188297
add a comment |Â
up vote
2
down vote
Yes. In general, if you have a sequence of $C^1$ functions $f_n$ on some interval and functions $f$ and $g$ such that $f_nto f$ pointwise and $f_n'to g$ uniformly, then $f'=g$. (Quick sketch of how to prove this without using integration: fix $x$ and $h$ and pick $n$ so that $f_n$ is sufficiently close to $f$ at $x$ and $x+h$ and $f_n'$ is sufficiently close to $g$ uniformly. Then $fracf(x+h)-f(x)h$ will be close to $fracf_n(x+h)-f_n(x)h$. By the mean value theorem the latter is equal to $f_n'(y)$ for some $y$ between $x+h$ and $x$, and $f_n'(y)$ is close to $g(y)$. Finally, if $h$ is sufficiently small, $g(y)$ will be close to $g(x)$ since $g$ is continuous.)
So, given $rinmathbbR$, pick a sequence of rational numbers $q_n$ converging to $r$ and let $f_n(x)=x^q_n$. Then $f_n(x)$ converges to $f(x)=x^r$ and $f_n'(x)=q_nx^q_n-1$ converges to $g(x)=rx^r-1$. Moreover, it is not hard to check that these convergences are uniform on any compact subset of $(0,infty)$. It thus follows that $f'=g$ on $(0,infty)$.
answered Jul 31 at 16:00
Eric Wofsey
161k12188297
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Yes. In general, if you have a sequence of $C^1$ functions $f_n$ on some interval and functions $f$ and $g$ such that $f_nto f$ pointwise and $f_n'to g$ uniformly, then $f'=g$. (Quick sketch of how to prove this without using integration: fix $x$ and $h$ and pick $n$ so that $f_n$ is sufficiently close to $f$ at $x$ and $x+h$ and $f_n'$ is sufficiently close to $g$ uniformly. Then $fracf(x+h)-f(x)h$ will be close to $fracf_n(x+h)-f_n(x)h$. By the mean value theorem the latter is equal to $f_n'(y)$ for some $y$ between $x+h$ and $x$, and $f_n'(y)$ is close to $g(y)$. Finally, if $h$ is sufficiently small, $g(y)$ will be close to $g(x)$ since $g$ is continuous.)
So, given $rinmathbbR$, pick a sequence of rational numbers $q_n$ converging to $r$ and let $f_n(x)=x^q_n$. Then $f_n(x)$ converges to $f(x)=x^r$ and $f_n'(x)=q_nx^q_n-1$ converges to $g(x)=rx^r-1$. Moreover, it is not hard to check that these convergences are uniform on any compact subset of $(0,infty)$. It thus follows that $f'=g$ on $(0,infty)$.
answered Jul 31 at 16:00
Eric Wofsey
161k12188297
Yes. In general, if you have a sequence of $C^1$ functions $f_n$ on some interval and functions $f$ and $g$ such that $f_nto f$ pointwise and $f_n'to g$ uniformly, then $f'=g$. (Quick sketch of how to prove this without using integration: fix $x$ and $h$ and pick $n$ so that $f_n$ is sufficiently close to $f$ at $x$ and $x+h$ and $f_n'$ is sufficiently close to $g$ uniformly. Then $fracf(x+h)-f(x)h$ will be close to $fracf_n(x+h)-f_n(x)h$. By the mean value theorem the latter is equal to $f_n'(y)$ for some $y$ between $x+h$ and $x$, and $f_n'(y)$ is close to $g(y)$. Finally, if $h$ is sufficiently small, $g(y)$ will be close to $g(x)$ since $g$ is continuous.)
So, given $rinmathbbR$, pick a sequence of rational numbers $q_n$ converging to $r$ and let $f_n(x)=x^q_n$. Then $f_n(x)$ converges to $f(x)=x^r$ and $f_n'(x)=q_nx^q_n-1$ converges to $g(x)=rx^r-1$. Moreover, it is not hard to check that these convergences are uniform on any compact subset of $(0,infty)$. It thus follows that $f'=g$ on $(0,infty)$.
answered Jul 31 at 16:00
Eric Wofsey
161k12188297
edited Jul 31 at 16:27
answered Jul 31 at 16:00
Eric Wofsey
161k12188297
answered Jul 31 at 16:00
Eric Wofsey
161k12188297
answered Jul 31 at 16:00
Eric Wofsey
161k12188297
161k12188297
add a comment |Â
add a comment |Â
up vote
0
down vote
Hint:
$$x^r:=lim_qto r,\qinmathbb Qx^q.$$
Then
$$(x^r)'=lim_hto0fraclim_qto r,\qinmathbb Q((x+h)^q-x^q)h=lim_qto r,\qinmathbb Qlim_hto0frac((x+h)^q-x^q)h=lim_qto r,\qinmathbb Qqx^q-1=rx^r-1.$$
The hard part is to justify the swap of the limits.
answered Jul 31 at 15:58
Yves Daoust
110k665203
add a comment |Â
up vote
0
down vote
Hint:
$$x^r:=lim_qto r,\qinmathbb Qx^q.$$
Then
$$(x^r)'=lim_hto0fraclim_qto r,\qinmathbb Q((x+h)^q-x^q)h=lim_qto r,\qinmathbb Qlim_hto0frac((x+h)^q-x^q)h=lim_qto r,\qinmathbb Qqx^q-1=rx^r-1.$$
The hard part is to justify the swap of the limits.
answered Jul 31 at 15:58
Yves Daoust
110k665203
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint:
$$x^r:=lim_qto r,\qinmathbb Qx^q.$$
Then
$$(x^r)'=lim_hto0fraclim_qto r,\qinmathbb Q((x+h)^q-x^q)h=lim_qto r,\qinmathbb Qlim_hto0frac((x+h)^q-x^q)h=lim_qto r,\qinmathbb Qqx^q-1=rx^r-1.$$
The hard part is to justify the swap of the limits.
answered Jul 31 at 15:58
Yves Daoust
110k665203
Hint:
$$x^r:=lim_qto r,\qinmathbb Qx^q.$$
Then
$$(x^r)'=lim_hto0fraclim_qto r,\qinmathbb Q((x+h)^q-x^q)h=lim_qto r,\qinmathbb Qlim_hto0frac((x+h)^q-x^q)h=lim_qto r,\qinmathbb Qqx^q-1=rx^r-1.$$
The hard part is to justify the swap of the limits.
answered Jul 31 at 15:58
Yves Daoust
110k665203
edited Jul 31 at 16:03
answered Jul 31 at 15:58
Yves Daoust
110k665203
answered Jul 31 at 15:58
Yves Daoust
110k665203
answered Jul 31 at 15:58
Yves Daoust
110k665203
110k665203
add a comment |Â
add a comment |Â
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Binomial theorem.
– copper.hat
Jul 31 at 15:29
1
@copper.hat Binomial theorem only allows you to derive this for $rin BbbN$
– Michal Dvořák
Jul 31 at 15:30
1
@MichalDvořák That’s not a duplicate at all, that’s about proving properties of the natural logarithm function, my question is about deriving the power rule without talking about $e$ and $ln(x)$ at all.
– Keshav Srinivasan
Jul 31 at 15:31
1
@MichalDvořák: The binomial theorem works for more than natural powers. It avoids using integrals which is what the OP was looking for.
– copper.hat
Jul 31 at 15:32
1
@ParamanandSingh Yes, you can. I gave the definition in my question: $x^r=lim_qrightarrow r x^q$ where $q$ is a variable that ranges over the rational numbers. I want to know how to use that definition to prove the power rule.
– Keshav Srinivasan
Jul 31 at 15:35