Mixing
Proof of product rule $(fg)^(n)$
Clash Royale CLAN TAG#URR8PPP
up vote
6
down vote
favorite
I went through induction proofs and they are nice. I'm just looking for an alternative.
To give some context, kindly consider the following example. To
expand $(x+a)^n$ we think of it as a combinatorics problem :
$$(x+a)(x+a)cdots (x+a)$$ To get $x$ term we need to
choose $x$ from any one product and $a$ from the rest. Thus the $x$
term would be $binomn1xa^n-1$
To get the $x^2$ term
we need to choose $x$ from any two products and $a$ from the rest:
$binomn2x^2a^n-2$
I'm wondering if the product rule can be seen using combinatorics $$beginalign
(fg)^' &=f'g+fg'\
(fg)^''&=(f'g+fg')^' = f''g+2f'g'+fg''
endalign$$
This looks almost same as the earlier problem of expanding $(x+a)^n$. I'm pretty sure these two problems are identical, but I'm not able to make the connection. Any help ?
calculus combinatorics derivatives
edited Jul 17 at 23:24
RayDansh
884215
asked Jul 17 at 23:08
rsadhvika
1,4891026
add a comment |Â
up vote
6
down vote
favorite
I went through induction proofs and they are nice. I'm just looking for an alternative.
To give some context, kindly consider the following example. To
expand $(x+a)^n$ we think of it as a combinatorics problem :
$$(x+a)(x+a)cdots (x+a)$$ To get $x$ term we need to
choose $x$ from any one product and $a$ from the rest. Thus the $x$
term would be $binomn1xa^n-1$
To get the $x^2$ term
we need to choose $x$ from any two products and $a$ from the rest:
$binomn2x^2a^n-2$
I'm wondering if the product rule can be seen using combinatorics $$beginalign
(fg)^' &=f'g+fg'\
(fg)^''&=(f'g+fg')^' = f''g+2f'g'+fg''
endalign$$
This looks almost same as the earlier problem of expanding $(x+a)^n$. I'm pretty sure these two problems are identical, but I'm not able to make the connection. Any help ?
calculus combinatorics derivatives
edited Jul 17 at 23:24
RayDansh
884215
asked Jul 17 at 23:08
rsadhvika
1,4891026
Did you see egreg's approach here?
– Cameron Buie
Jul 17 at 23:24
Yes @CameronBuie I'm familiar with that proof and I like induction a lot. I'm trying to interpret the product rule using arguments like choose two first derivatives, choose three second derivatives, etc..
– rsadhvika
Jul 17 at 23:30
I mean only after I wrote $(x+a)^n$ flat $$(x+a)(x+a)cdots (x+a)$$ it became clear what I had to choose and leave to get the coefficient of $x^k$. I'm wondering if we can flatten something similar in product rule.. @CameronBuie
– rsadhvika
Jul 17 at 23:36
The required connection is done in the Species Theory. For every two species $F$ and $G$, $[F.G]' = F'G + G'F$. Here is the derivative of a product math.stackexchange.com/questions/2852683/…
– nbaxter
Jul 20 at 18:39
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I went through induction proofs and they are nice. I'm just looking for an alternative.
To give some context, kindly consider the following example. To
expand $(x+a)^n$ we think of it as a combinatorics problem :
$$(x+a)(x+a)cdots (x+a)$$ To get $x$ term we need to
choose $x$ from any one product and $a$ from the rest. Thus the $x$
term would be $binomn1xa^n-1$
To get the $x^2$ term
we need to choose $x$ from any two products and $a$ from the rest:
$binomn2x^2a^n-2$
I'm wondering if the product rule can be seen using combinatorics $$beginalign
(fg)^' &=f'g+fg'\
(fg)^''&=(f'g+fg')^' = f''g+2f'g'+fg''
endalign$$
This looks almost same as the earlier problem of expanding $(x+a)^n$. I'm pretty sure these two problems are identical, but I'm not able to make the connection. Any help ?
calculus combinatorics derivatives
edited Jul 17 at 23:24
RayDansh
884215
asked Jul 17 at 23:08
rsadhvika
1,4891026
I went through induction proofs and they are nice. I'm just looking for an alternative.
To give some context, kindly consider the following example. To
expand $(x+a)^n$ we think of it as a combinatorics problem :
$$(x+a)(x+a)cdots (x+a)$$ To get $x$ term we need to
choose $x$ from any one product and $a$ from the rest. Thus the $x$
term would be $binomn1xa^n-1$
To get the $x^2$ term
we need to choose $x$ from any two products and $a$ from the rest:
$binomn2x^2a^n-2$
I'm wondering if the product rule can be seen using combinatorics $$beginalign
(fg)^' &=f'g+fg'\
(fg)^''&=(f'g+fg')^' = f''g+2f'g'+fg''
endalign$$
This looks almost same as the earlier problem of expanding $(x+a)^n$. I'm pretty sure these two problems are identical, but I'm not able to make the connection. Any help ?
calculus combinatorics derivatives
edited Jul 17 at 23:24
RayDansh
884215
asked Jul 17 at 23:08
rsadhvika
1,4891026
edited Jul 17 at 23:24
RayDansh
884215
edited Jul 17 at 23:24
RayDansh
884215
edited Jul 17 at 23:24
RayDansh
884215
884215
asked Jul 17 at 23:08
rsadhvika
1,4891026
asked Jul 17 at 23:08
rsadhvika
1,4891026
asked Jul 17 at 23:08
rsadhvika
1,4891026
1,4891026
Did you see egreg's approach here?
– Cameron Buie
Jul 17 at 23:24
Yes @CameronBuie I'm familiar with that proof and I like induction a lot. I'm trying to interpret the product rule using arguments like choose two first derivatives, choose three second derivatives, etc..
– rsadhvika
Jul 17 at 23:30
I mean only after I wrote $(x+a)^n$ flat $$(x+a)(x+a)cdots (x+a)$$ it became clear what I had to choose and leave to get the coefficient of $x^k$. I'm wondering if we can flatten something similar in product rule.. @CameronBuie
– rsadhvika
Jul 17 at 23:36
The required connection is done in the Species Theory. For every two species $F$ and $G$, $[F.G]' = F'G + G'F$. Here is the derivative of a product math.stackexchange.com/questions/2852683/…
– nbaxter
Jul 20 at 18:39
add a comment |Â
Did you see egreg's approach here?
– Cameron Buie
Jul 17 at 23:24
Yes @CameronBuie I'm familiar with that proof and I like induction a lot. I'm trying to interpret the product rule using arguments like choose two first derivatives, choose three second derivatives, etc..
– rsadhvika
Jul 17 at 23:30
I mean only after I wrote $(x+a)^n$ flat $$(x+a)(x+a)cdots (x+a)$$ it became clear what I had to choose and leave to get the coefficient of $x^k$. I'm wondering if we can flatten something similar in product rule.. @CameronBuie
– rsadhvika
Jul 17 at 23:36
The required connection is done in the Species Theory. For every two species $F$ and $G$, $[F.G]' = F'G + G'F$. Here is the derivative of a product math.stackexchange.com/questions/2852683/…
– nbaxter
Jul 20 at 18:39
Did you see egreg's approach here?
– Cameron Buie
Jul 17 at 23:24
Did you see egreg's approach here?
– Cameron Buie
Jul 17 at 23:24
Yes @CameronBuie I'm familiar with that proof and I like induction a lot. I'm trying to interpret the product rule using arguments like choose two first derivatives, choose three second derivatives, etc..
– rsadhvika
Jul 17 at 23:30
Yes @CameronBuie I'm familiar with that proof and I like induction a lot. I'm trying to interpret the product rule using arguments like choose two first derivatives, choose three second derivatives, etc..
– rsadhvika
Jul 17 at 23:30
I mean only after I wrote $(x+a)^n$ flat $$(x+a)(x+a)cdots (x+a)$$ it became clear what I had to choose and leave to get the coefficient of $x^k$. I'm wondering if we can flatten something similar in product rule.. @CameronBuie
– rsadhvika
Jul 17 at 23:36
I mean only after I wrote $(x+a)^n$ flat $$(x+a)(x+a)cdots (x+a)$$ it became clear what I had to choose and leave to get the coefficient of $x^k$. I'm wondering if we can flatten something similar in product rule.. @CameronBuie
– rsadhvika
Jul 17 at 23:36
The required connection is done in the Species Theory. For every two species $F$ and $G$, $[F.G]' = F'G + G'F$. Here is the derivative of a product math.stackexchange.com/questions/2852683/…
– nbaxter
Jul 20 at 18:39
The required connection is done in the Species Theory. For every two species $F$ and $G$, $[F.G]' = F'G + G'F$. Here is the derivative of a product math.stackexchange.com/questions/2852683/…
– nbaxter
Jul 20 at 18:39
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
Define three operators: $D_times$ represents differentiation of a product; $D_1$ represents differentiation of the first term of the product; and $D_2$ represents differentiation of the second term. Then the simple product rule $(fg)' = f'g + fg'$ can be written $D_times = D_1 + D_2$. Observe that $D_1$ and $D_2$ are commutative, so you can apply the binomial theorem to $(D_1 + D_2)^n$.
answered Jul 18 at 10:14
Peter Taylor
7,77512239
add a comment |Â
up vote
5
down vote
A "counting argument" could be as follows. Let $mu$ be the multiplication operator and let $D$ be the derivative operator. Then the usual product rule says that we have the following identity.
$$Dmu = mu(Dtimes 1+1times D)$$
This means, by induction, that
$$D^nmu = mu(Dtimes 1+1times D)^n$$
Now expand the right hand side using the binomial theorem! This is why both proofs are the same.
Induction, of course. Suppose that it is true that
$$D^n(fcdot g) = sum_i=0^n binom ni D^ifcdot D^n-i g$$
Applying $D$ you get
$$D^n+1(fcdot g) = sum_i=0^n binom ni D(D^ifcdot D^n-i g)$$
Apply the base case to this to get
$$D^n+1(fcdot g) = sum_i=0^n binom ni D^i+1fcdot D^n-i g+D^ifcdot D^n-i+1 g.$$
Now use Pascal's rule to obtain the induction step. Note the proof is the same as for the binomial.
answered Jul 17 at 23:14
Pedro Tamaroff♦
93.8k10143290
3
Though, OP did mention that he was looking to an alternative to induction.
– Argon
Jul 17 at 23:17
Yeah I went through many induction proofs in another post yesterday, but this form using operators is really slick. Thank you :) I'm still wondering if we can approach the problem directly with out referring to induction explicitly. Like using arguments like choosing terms etc.. Hmm
– rsadhvika
Jul 17 at 23:21
add a comment |Â
up vote
2
down vote
With (as usual) $f^(n)$ denoting the $n$th derivative of $f$ when $n>0,$ and $f^(0)=f.$
There are $n$ steps ($n$ differentiations ) to get from $(fg)^(0)$ to $(fg)^(n).$
After the $m$th step ($mgeq 0$) we have the sum of a finite sequence of (not necessarily unequal) terms, each of the form $f^(j)g^(m-j)$ for some $0leq jleq m.$ The $(m+1)$th step replaces each such term with the 2 terms $f^(j+1)g^(m-j)$ and $f^(j)g^(m+1-j).$
After $n$ steps a term $f^(j)g^(n-j)$ will appear $binom nj$ times because there are $binom nj$ different "paths" through the $n$ steps that will result in such a term.
answered Jul 18 at 1:09
DanielWainfleet
31.7k31643
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Define three operators: $D_times$ represents differentiation of a product; $D_1$ represents differentiation of the first term of the product; and $D_2$ represents differentiation of the second term. Then the simple product rule $(fg)' = f'g + fg'$ can be written $D_times = D_1 + D_2$. Observe that $D_1$ and $D_2$ are commutative, so you can apply the binomial theorem to $(D_1 + D_2)^n$.
answered Jul 18 at 10:14
Peter Taylor
7,77512239
add a comment |Â
up vote
3
down vote
accepted
Define three operators: $D_times$ represents differentiation of a product; $D_1$ represents differentiation of the first term of the product; and $D_2$ represents differentiation of the second term. Then the simple product rule $(fg)' = f'g + fg'$ can be written $D_times = D_1 + D_2$. Observe that $D_1$ and $D_2$ are commutative, so you can apply the binomial theorem to $(D_1 + D_2)^n$.
answered Jul 18 at 10:14
Peter Taylor
7,77512239
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Define three operators: $D_times$ represents differentiation of a product; $D_1$ represents differentiation of the first term of the product; and $D_2$ represents differentiation of the second term. Then the simple product rule $(fg)' = f'g + fg'$ can be written $D_times = D_1 + D_2$. Observe that $D_1$ and $D_2$ are commutative, so you can apply the binomial theorem to $(D_1 + D_2)^n$.
answered Jul 18 at 10:14
Peter Taylor
7,77512239
Define three operators: $D_times$ represents differentiation of a product; $D_1$ represents differentiation of the first term of the product; and $D_2$ represents differentiation of the second term. Then the simple product rule $(fg)' = f'g + fg'$ can be written $D_times = D_1 + D_2$. Observe that $D_1$ and $D_2$ are commutative, so you can apply the binomial theorem to $(D_1 + D_2)^n$.
answered Jul 18 at 10:14
Peter Taylor
7,77512239
answered Jul 18 at 10:14
Peter Taylor
7,77512239
answered Jul 18 at 10:14
Peter Taylor
7,77512239
answered Jul 18 at 10:14
Peter Taylor
7,77512239
7,77512239
add a comment |Â
add a comment |Â
up vote
5
down vote
A "counting argument" could be as follows. Let $mu$ be the multiplication operator and let $D$ be the derivative operator. Then the usual product rule says that we have the following identity.
$$Dmu = mu(Dtimes 1+1times D)$$
This means, by induction, that
$$D^nmu = mu(Dtimes 1+1times D)^n$$
Now expand the right hand side using the binomial theorem! This is why both proofs are the same.
Induction, of course. Suppose that it is true that
$$D^n(fcdot g) = sum_i=0^n binom ni D^ifcdot D^n-i g$$
Applying $D$ you get
$$D^n+1(fcdot g) = sum_i=0^n binom ni D(D^ifcdot D^n-i g)$$
Apply the base case to this to get
$$D^n+1(fcdot g) = sum_i=0^n binom ni D^i+1fcdot D^n-i g+D^ifcdot D^n-i+1 g.$$
Now use Pascal's rule to obtain the induction step. Note the proof is the same as for the binomial.
answered Jul 17 at 23:14
Pedro Tamaroff♦
93.8k10143290
3
Though, OP did mention that he was looking to an alternative to induction.
– Argon
Jul 17 at 23:17
Yeah I went through many induction proofs in another post yesterday, but this form using operators is really slick. Thank you :) I'm still wondering if we can approach the problem directly with out referring to induction explicitly. Like using arguments like choosing terms etc.. Hmm
– rsadhvika
Jul 17 at 23:21
add a comment |Â
up vote
5
down vote
A "counting argument" could be as follows. Let $mu$ be the multiplication operator and let $D$ be the derivative operator. Then the usual product rule says that we have the following identity.
$$Dmu = mu(Dtimes 1+1times D)$$
This means, by induction, that
$$D^nmu = mu(Dtimes 1+1times D)^n$$
Now expand the right hand side using the binomial theorem! This is why both proofs are the same.
Induction, of course. Suppose that it is true that
$$D^n(fcdot g) = sum_i=0^n binom ni D^ifcdot D^n-i g$$
Applying $D$ you get
$$D^n+1(fcdot g) = sum_i=0^n binom ni D(D^ifcdot D^n-i g)$$
Apply the base case to this to get
$$D^n+1(fcdot g) = sum_i=0^n binom ni D^i+1fcdot D^n-i g+D^ifcdot D^n-i+1 g.$$
Now use Pascal's rule to obtain the induction step. Note the proof is the same as for the binomial.
answered Jul 17 at 23:14
Pedro Tamaroff♦
93.8k10143290
3
Though, OP did mention that he was looking to an alternative to induction.
– Argon
Jul 17 at 23:17
Yeah I went through many induction proofs in another post yesterday, but this form using operators is really slick. Thank you :) I'm still wondering if we can approach the problem directly with out referring to induction explicitly. Like using arguments like choosing terms etc.. Hmm
– rsadhvika
Jul 17 at 23:21
add a comment |Â
up vote
5
down vote
up vote
5
down vote
A "counting argument" could be as follows. Let $mu$ be the multiplication operator and let $D$ be the derivative operator. Then the usual product rule says that we have the following identity.
$$Dmu = mu(Dtimes 1+1times D)$$
This means, by induction, that
$$D^nmu = mu(Dtimes 1+1times D)^n$$
Now expand the right hand side using the binomial theorem! This is why both proofs are the same.
Induction, of course. Suppose that it is true that
$$D^n(fcdot g) = sum_i=0^n binom ni D^ifcdot D^n-i g$$
Applying $D$ you get
$$D^n+1(fcdot g) = sum_i=0^n binom ni D(D^ifcdot D^n-i g)$$
Apply the base case to this to get
$$D^n+1(fcdot g) = sum_i=0^n binom ni D^i+1fcdot D^n-i g+D^ifcdot D^n-i+1 g.$$
Now use Pascal's rule to obtain the induction step. Note the proof is the same as for the binomial.
answered Jul 17 at 23:14
Pedro Tamaroff♦
93.8k10143290
A "counting argument" could be as follows. Let $mu$ be the multiplication operator and let $D$ be the derivative operator. Then the usual product rule says that we have the following identity.
$$Dmu = mu(Dtimes 1+1times D)$$
This means, by induction, that
$$D^nmu = mu(Dtimes 1+1times D)^n$$
Now expand the right hand side using the binomial theorem! This is why both proofs are the same.
Induction, of course. Suppose that it is true that
$$D^n(fcdot g) = sum_i=0^n binom ni D^ifcdot D^n-i g$$
Applying $D$ you get
$$D^n+1(fcdot g) = sum_i=0^n binom ni D(D^ifcdot D^n-i g)$$
Apply the base case to this to get
$$D^n+1(fcdot g) = sum_i=0^n binom ni D^i+1fcdot D^n-i g+D^ifcdot D^n-i+1 g.$$
Now use Pascal's rule to obtain the induction step. Note the proof is the same as for the binomial.
answered Jul 17 at 23:14
Pedro Tamaroff♦
93.8k10143290
edited Jul 17 at 23:34
answered Jul 17 at 23:14
Pedro Tamaroff♦
93.8k10143290
answered Jul 17 at 23:14
Pedro Tamaroff♦
93.8k10143290
answered Jul 17 at 23:14
Pedro Tamaroff♦
93.8k10143290
93.8k10143290
3
Though, OP did mention that he was looking to an alternative to induction.
– Argon
Jul 17 at 23:17
Yeah I went through many induction proofs in another post yesterday, but this form using operators is really slick. Thank you :) I'm still wondering if we can approach the problem directly with out referring to induction explicitly. Like using arguments like choosing terms etc.. Hmm
– rsadhvika
Jul 17 at 23:21
add a comment |Â
3
Though, OP did mention that he was looking to an alternative to induction.
– Argon
Jul 17 at 23:17
Yeah I went through many induction proofs in another post yesterday, but this form using operators is really slick. Thank you :) I'm still wondering if we can approach the problem directly with out referring to induction explicitly. Like using arguments like choosing terms etc.. Hmm
– rsadhvika
Jul 17 at 23:21
3
3
Though, OP did mention that he was looking to an alternative to induction.
– Argon
Jul 17 at 23:17
Though, OP did mention that he was looking to an alternative to induction.
– Argon
Jul 17 at 23:17
Yeah I went through many induction proofs in another post yesterday, but this form using operators is really slick. Thank you :) I'm still wondering if we can approach the problem directly with out referring to induction explicitly. Like using arguments like choosing terms etc.. Hmm
– rsadhvika
Jul 17 at 23:21
Yeah I went through many induction proofs in another post yesterday, but this form using operators is really slick. Thank you :) I'm still wondering if we can approach the problem directly with out referring to induction explicitly. Like using arguments like choosing terms etc.. Hmm
– rsadhvika
Jul 17 at 23:21
add a comment |Â
up vote
2
down vote
With (as usual) $f^(n)$ denoting the $n$th derivative of $f$ when $n>0,$ and $f^(0)=f.$
There are $n$ steps ($n$ differentiations ) to get from $(fg)^(0)$ to $(fg)^(n).$
After the $m$th step ($mgeq 0$) we have the sum of a finite sequence of (not necessarily unequal) terms, each of the form $f^(j)g^(m-j)$ for some $0leq jleq m.$ The $(m+1)$th step replaces each such term with the 2 terms $f^(j+1)g^(m-j)$ and $f^(j)g^(m+1-j).$
After $n$ steps a term $f^(j)g^(n-j)$ will appear $binom nj$ times because there are $binom nj$ different "paths" through the $n$ steps that will result in such a term.
answered Jul 18 at 1:09
DanielWainfleet
31.7k31643
add a comment |Â
up vote
2
down vote
With (as usual) $f^(n)$ denoting the $n$th derivative of $f$ when $n>0,$ and $f^(0)=f.$
There are $n$ steps ($n$ differentiations ) to get from $(fg)^(0)$ to $(fg)^(n).$
After the $m$th step ($mgeq 0$) we have the sum of a finite sequence of (not necessarily unequal) terms, each of the form $f^(j)g^(m-j)$ for some $0leq jleq m.$ The $(m+1)$th step replaces each such term with the 2 terms $f^(j+1)g^(m-j)$ and $f^(j)g^(m+1-j).$
After $n$ steps a term $f^(j)g^(n-j)$ will appear $binom nj$ times because there are $binom nj$ different "paths" through the $n$ steps that will result in such a term.
answered Jul 18 at 1:09
DanielWainfleet
31.7k31643
add a comment |Â
up vote
2
down vote
up vote
2
down vote
With (as usual) $f^(n)$ denoting the $n$th derivative of $f$ when $n>0,$ and $f^(0)=f.$
There are $n$ steps ($n$ differentiations ) to get from $(fg)^(0)$ to $(fg)^(n).$
After the $m$th step ($mgeq 0$) we have the sum of a finite sequence of (not necessarily unequal) terms, each of the form $f^(j)g^(m-j)$ for some $0leq jleq m.$ The $(m+1)$th step replaces each such term with the 2 terms $f^(j+1)g^(m-j)$ and $f^(j)g^(m+1-j).$
After $n$ steps a term $f^(j)g^(n-j)$ will appear $binom nj$ times because there are $binom nj$ different "paths" through the $n$ steps that will result in such a term.
answered Jul 18 at 1:09
DanielWainfleet
31.7k31643
With (as usual) $f^(n)$ denoting the $n$th derivative of $f$ when $n>0,$ and $f^(0)=f.$
There are $n$ steps ($n$ differentiations ) to get from $(fg)^(0)$ to $(fg)^(n).$
After the $m$th step ($mgeq 0$) we have the sum of a finite sequence of (not necessarily unequal) terms, each of the form $f^(j)g^(m-j)$ for some $0leq jleq m.$ The $(m+1)$th step replaces each such term with the 2 terms $f^(j+1)g^(m-j)$ and $f^(j)g^(m+1-j).$
After $n$ steps a term $f^(j)g^(n-j)$ will appear $binom nj$ times because there are $binom nj$ different "paths" through the $n$ steps that will result in such a term.
answered Jul 18 at 1:09
DanielWainfleet
31.7k31643
answered Jul 18 at 1:09
DanielWainfleet
31.7k31643
answered Jul 18 at 1:09
DanielWainfleet
31.7k31643
answered Jul 18 at 1:09
DanielWainfleet
31.7k31643
31.7k31643
add a comment |Â
add a comment |Â
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Did you see egreg's approach here?
– Cameron Buie
Jul 17 at 23:24
Yes @CameronBuie I'm familiar with that proof and I like induction a lot. I'm trying to interpret the product rule using arguments like choose two first derivatives, choose three second derivatives, etc..
– rsadhvika
Jul 17 at 23:30
I mean only after I wrote $(x+a)^n$ flat $$(x+a)(x+a)cdots (x+a)$$ it became clear what I had to choose and leave to get the coefficient of $x^k$. I'm wondering if we can flatten something similar in product rule.. @CameronBuie
– rsadhvika
Jul 17 at 23:36
The required connection is done in the Species Theory. For every two species $F$ and $G$, $[F.G]' = F'G + G'F$. Here is the derivative of a product math.stackexchange.com/questions/2852683/…
– nbaxter
Jul 20 at 18:39