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Induction differential proof
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So I was doing an induction proof that contained a differential. Now I got through most of it but lastly in order to complete the proof I needed to prove
$$fracd^k dx^kx^k=k!$$
Now I don't need to do this via induction or anything it's just that inside the actual induction proof this statement was there and in order to prove the $(k+1)$-th term I need to show this.
I can see how this would happen:
$fracd dxx^k = k(x^k-1)$
$fracd^2 dx^2x^k = k(k-1)x^k-2$
..
...
..
Until eventually I would get
$kcdot (k-1)cdot (k-2)cdot (k-3)cdot ... cdot x^k-k$
Which in turn would be $k!$
However is there any better way of showing this, like a more formal proof or better strucutred one ?
derivatives proof-writing induction
edited 7 hours ago
Robert Howard
1,241620
asked Aug 3 at 22:12
user122343
445
add a comment |Â
up vote
4
down vote
favorite
So I was doing an induction proof that contained a differential. Now I got through most of it but lastly in order to complete the proof I needed to prove
$$fracd^k dx^kx^k=k!$$
Now I don't need to do this via induction or anything it's just that inside the actual induction proof this statement was there and in order to prove the $(k+1)$-th term I need to show this.
I can see how this would happen:
$fracd dxx^k = k(x^k-1)$
$fracd^2 dx^2x^k = k(k-1)x^k-2$
..
...
..
Until eventually I would get
$kcdot (k-1)cdot (k-2)cdot (k-3)cdot ... cdot x^k-k$
Which in turn would be $k!$
However is there any better way of showing this, like a more formal proof or better strucutred one ?
derivatives proof-writing induction
edited 7 hours ago
Robert Howard
1,241620
asked Aug 3 at 22:12
user122343
445
Its not hard to prove it with induction.
– Calvin Khor
Aug 3 at 22:18
What you wrote looks good enough if only it was typeset in MathJax.
– Arnaud Mortier
Aug 3 at 22:20
Besides proving $fracd^kdx^k(x^k) = k!$ by induction on $k$, you could also formalize your argument directly: prove by induction on $n$ that whenever $1 le n le k$, then $fracd^ndx^n(x^k) = frack!(k-n)! x^k-n$.
– Daniel Schepler
Aug 3 at 22:25
A tutorial for typing mathematics in MathJax can be found here, by the way: math.meta.stackexchange.com/questions/5020/…
– Robert Howard
Aug 3 at 22:25
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
So I was doing an induction proof that contained a differential. Now I got through most of it but lastly in order to complete the proof I needed to prove
$$fracd^k dx^kx^k=k!$$
Now I don't need to do this via induction or anything it's just that inside the actual induction proof this statement was there and in order to prove the $(k+1)$-th term I need to show this.
I can see how this would happen:
$fracd dxx^k = k(x^k-1)$
$fracd^2 dx^2x^k = k(k-1)x^k-2$
..
...
..
Until eventually I would get
$kcdot (k-1)cdot (k-2)cdot (k-3)cdot ... cdot x^k-k$
Which in turn would be $k!$
However is there any better way of showing this, like a more formal proof or better strucutred one ?
derivatives proof-writing induction
edited 7 hours ago
Robert Howard
1,241620
asked Aug 3 at 22:12
user122343
445
So I was doing an induction proof that contained a differential. Now I got through most of it but lastly in order to complete the proof I needed to prove
$$fracd^k dx^kx^k=k!$$
Now I don't need to do this via induction or anything it's just that inside the actual induction proof this statement was there and in order to prove the $(k+1)$-th term I need to show this.
I can see how this would happen:
$fracd dxx^k = k(x^k-1)$
$fracd^2 dx^2x^k = k(k-1)x^k-2$
..
...
..
Until eventually I would get
$kcdot (k-1)cdot (k-2)cdot (k-3)cdot ... cdot x^k-k$
Which in turn would be $k!$
However is there any better way of showing this, like a more formal proof or better strucutred one ?
derivatives proof-writing induction
edited 7 hours ago
Robert Howard
1,241620
asked Aug 3 at 22:12
user122343
445
edited 7 hours ago
Robert Howard
1,241620
edited 7 hours ago
Robert Howard
1,241620
edited 7 hours ago
Robert Howard
1,241620
1,241620
asked Aug 3 at 22:12
user122343
445
asked Aug 3 at 22:12
user122343
445
asked Aug 3 at 22:12
user122343
445
445
Its not hard to prove it with induction.
– Calvin Khor
Aug 3 at 22:18
What you wrote looks good enough if only it was typeset in MathJax.
– Arnaud Mortier
Aug 3 at 22:20
Besides proving $fracd^kdx^k(x^k) = k!$ by induction on $k$, you could also formalize your argument directly: prove by induction on $n$ that whenever $1 le n le k$, then $fracd^ndx^n(x^k) = frack!(k-n)! x^k-n$.
– Daniel Schepler
Aug 3 at 22:25
A tutorial for typing mathematics in MathJax can be found here, by the way: math.meta.stackexchange.com/questions/5020/…
– Robert Howard
Aug 3 at 22:25
add a comment |Â
Its not hard to prove it with induction.
– Calvin Khor
Aug 3 at 22:18
What you wrote looks good enough if only it was typeset in MathJax.
– Arnaud Mortier
Aug 3 at 22:20
Besides proving $fracd^kdx^k(x^k) = k!$ by induction on $k$, you could also formalize your argument directly: prove by induction on $n$ that whenever $1 le n le k$, then $fracd^ndx^n(x^k) = frack!(k-n)! x^k-n$.
– Daniel Schepler
Aug 3 at 22:25
A tutorial for typing mathematics in MathJax can be found here, by the way: math.meta.stackexchange.com/questions/5020/…
– Robert Howard
Aug 3 at 22:25
Its not hard to prove it with induction.
– Calvin Khor
Aug 3 at 22:18
Its not hard to prove it with induction.
– Calvin Khor
Aug 3 at 22:18
What you wrote looks good enough if only it was typeset in MathJax.
– Arnaud Mortier
Aug 3 at 22:20
What you wrote looks good enough if only it was typeset in MathJax.
– Arnaud Mortier
Aug 3 at 22:20
Besides proving $fracd^kdx^k(x^k) = k!$ by induction on $k$, you could also formalize your argument directly: prove by induction on $n$ that whenever $1 le n le k$, then $fracd^ndx^n(x^k) = frack!(k-n)! x^k-n$.
– Daniel Schepler
Aug 3 at 22:25
Besides proving $fracd^kdx^k(x^k) = k!$ by induction on $k$, you could also formalize your argument directly: prove by induction on $n$ that whenever $1 le n le k$, then $fracd^ndx^n(x^k) = frack!(k-n)! x^k-n$.
– Daniel Schepler
Aug 3 at 22:25
A tutorial for typing mathematics in MathJax can be found here, by the way: math.meta.stackexchange.com/questions/5020/…
– Robert Howard
Aug 3 at 22:25
A tutorial for typing mathematics in MathJax can be found here, by the way: math.meta.stackexchange.com/questions/5020/…
– Robert Howard
Aug 3 at 22:25
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
You wrote that you don't need to do this via induction, but it's a good way of doing it. What I mean by this is that you can prove by induction on $k$ that $fracmathrm d^kmathrm dx^kx^n=n(n-1)cdots(n-k+1)x^n-k$, where $ngeqslant k$. If $k=1$, this is just the assertion that $fracmathrm dmathrm dxx=1$. Suppose that it is true for some $k$. Thenbeginalignfracmathrm d^k+1mathrm dx^k+1x^n&=fracmathrm dmathrm dxleft(fracmathrm d^kmathrm dx^kx^nright)\&=fracmathrm dmathrm dxleft(n(n-1)cdots(n-k+1)x^n-kright)\&=n(n-1)cdots(n-k+1)(n-k)x^n-k-1endalign
answered Aug 3 at 22:25
José Carlos Santos
112k1696172
You go exactly the opposite way with your induction than I do, José :-)
– Nicolas FRANCOIS
Aug 3 at 22:27
add a comment |Â
up vote
6
down vote
Another way : suppose $fracd^ndx^n(x^n) = n!$ for some $ninmathbf N$. Then :
$$fracd^n+1dx^n+1(x^n+1) = fracd^ndx^nleft(fracddx(x^n+1)right) = fracd^ndx^n((n+1)x^n) = (n+1)fracd^ndx^n(x^n) = (n+1)n! = (n+1)!$$
There you have to know the first derivative of $x^n$ for any $n$, which can be proved... by induction, knowing the derivative of a product and the derivative of $x$ and $1$ :-)
answered Aug 3 at 22:26
Nicolas FRANCOIS
3,1121414
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
accepted
You wrote that you don't need to do this via induction, but it's a good way of doing it. What I mean by this is that you can prove by induction on $k$ that $fracmathrm d^kmathrm dx^kx^n=n(n-1)cdots(n-k+1)x^n-k$, where $ngeqslant k$. If $k=1$, this is just the assertion that $fracmathrm dmathrm dxx=1$. Suppose that it is true for some $k$. Thenbeginalignfracmathrm d^k+1mathrm dx^k+1x^n&=fracmathrm dmathrm dxleft(fracmathrm d^kmathrm dx^kx^nright)\&=fracmathrm dmathrm dxleft(n(n-1)cdots(n-k+1)x^n-kright)\&=n(n-1)cdots(n-k+1)(n-k)x^n-k-1endalign
answered Aug 3 at 22:25
José Carlos Santos
112k1696172
You go exactly the opposite way with your induction than I do, José :-)
– Nicolas FRANCOIS
Aug 3 at 22:27
add a comment |Â
up vote
2
down vote
accepted
You wrote that you don't need to do this via induction, but it's a good way of doing it. What I mean by this is that you can prove by induction on $k$ that $fracmathrm d^kmathrm dx^kx^n=n(n-1)cdots(n-k+1)x^n-k$, where $ngeqslant k$. If $k=1$, this is just the assertion that $fracmathrm dmathrm dxx=1$. Suppose that it is true for some $k$. Thenbeginalignfracmathrm d^k+1mathrm dx^k+1x^n&=fracmathrm dmathrm dxleft(fracmathrm d^kmathrm dx^kx^nright)\&=fracmathrm dmathrm dxleft(n(n-1)cdots(n-k+1)x^n-kright)\&=n(n-1)cdots(n-k+1)(n-k)x^n-k-1endalign
answered Aug 3 at 22:25
José Carlos Santos
112k1696172
You go exactly the opposite way with your induction than I do, José :-)
– Nicolas FRANCOIS
Aug 3 at 22:27
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You wrote that you don't need to do this via induction, but it's a good way of doing it. What I mean by this is that you can prove by induction on $k$ that $fracmathrm d^kmathrm dx^kx^n=n(n-1)cdots(n-k+1)x^n-k$, where $ngeqslant k$. If $k=1$, this is just the assertion that $fracmathrm dmathrm dxx=1$. Suppose that it is true for some $k$. Thenbeginalignfracmathrm d^k+1mathrm dx^k+1x^n&=fracmathrm dmathrm dxleft(fracmathrm d^kmathrm dx^kx^nright)\&=fracmathrm dmathrm dxleft(n(n-1)cdots(n-k+1)x^n-kright)\&=n(n-1)cdots(n-k+1)(n-k)x^n-k-1endalign
answered Aug 3 at 22:25
José Carlos Santos
112k1696172
You wrote that you don't need to do this via induction, but it's a good way of doing it. What I mean by this is that you can prove by induction on $k$ that $fracmathrm d^kmathrm dx^kx^n=n(n-1)cdots(n-k+1)x^n-k$, where $ngeqslant k$. If $k=1$, this is just the assertion that $fracmathrm dmathrm dxx=1$. Suppose that it is true for some $k$. Thenbeginalignfracmathrm d^k+1mathrm dx^k+1x^n&=fracmathrm dmathrm dxleft(fracmathrm d^kmathrm dx^kx^nright)\&=fracmathrm dmathrm dxleft(n(n-1)cdots(n-k+1)x^n-kright)\&=n(n-1)cdots(n-k+1)(n-k)x^n-k-1endalign
answered Aug 3 at 22:25
José Carlos Santos
112k1696172
answered Aug 3 at 22:25
José Carlos Santos
112k1696172
answered Aug 3 at 22:25
José Carlos Santos
112k1696172
answered Aug 3 at 22:25
José Carlos Santos
112k1696172
112k1696172
You go exactly the opposite way with your induction than I do, José :-)
– Nicolas FRANCOIS
Aug 3 at 22:27
add a comment |Â
You go exactly the opposite way with your induction than I do, José :-)
– Nicolas FRANCOIS
Aug 3 at 22:27
You go exactly the opposite way with your induction than I do, José :-)
– Nicolas FRANCOIS
Aug 3 at 22:27
You go exactly the opposite way with your induction than I do, José :-)
– Nicolas FRANCOIS
Aug 3 at 22:27
add a comment |Â
up vote
6
down vote
Another way : suppose $fracd^ndx^n(x^n) = n!$ for some $ninmathbf N$. Then :
$$fracd^n+1dx^n+1(x^n+1) = fracd^ndx^nleft(fracddx(x^n+1)right) = fracd^ndx^n((n+1)x^n) = (n+1)fracd^ndx^n(x^n) = (n+1)n! = (n+1)!$$
There you have to know the first derivative of $x^n$ for any $n$, which can be proved... by induction, knowing the derivative of a product and the derivative of $x$ and $1$ :-)
answered Aug 3 at 22:26
Nicolas FRANCOIS
3,1121414
add a comment |Â
up vote
6
down vote
Another way : suppose $fracd^ndx^n(x^n) = n!$ for some $ninmathbf N$. Then :
$$fracd^n+1dx^n+1(x^n+1) = fracd^ndx^nleft(fracddx(x^n+1)right) = fracd^ndx^n((n+1)x^n) = (n+1)fracd^ndx^n(x^n) = (n+1)n! = (n+1)!$$
There you have to know the first derivative of $x^n$ for any $n$, which can be proved... by induction, knowing the derivative of a product and the derivative of $x$ and $1$ :-)
answered Aug 3 at 22:26
Nicolas FRANCOIS
3,1121414
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Another way : suppose $fracd^ndx^n(x^n) = n!$ for some $ninmathbf N$. Then :
$$fracd^n+1dx^n+1(x^n+1) = fracd^ndx^nleft(fracddx(x^n+1)right) = fracd^ndx^n((n+1)x^n) = (n+1)fracd^ndx^n(x^n) = (n+1)n! = (n+1)!$$
There you have to know the first derivative of $x^n$ for any $n$, which can be proved... by induction, knowing the derivative of a product and the derivative of $x$ and $1$ :-)
answered Aug 3 at 22:26
Nicolas FRANCOIS
3,1121414
Another way : suppose $fracd^ndx^n(x^n) = n!$ for some $ninmathbf N$. Then :
$$fracd^n+1dx^n+1(x^n+1) = fracd^ndx^nleft(fracddx(x^n+1)right) = fracd^ndx^n((n+1)x^n) = (n+1)fracd^ndx^n(x^n) = (n+1)n! = (n+1)!$$
There you have to know the first derivative of $x^n$ for any $n$, which can be proved... by induction, knowing the derivative of a product and the derivative of $x$ and $1$ :-)
answered Aug 3 at 22:26
Nicolas FRANCOIS
3,1121414
answered Aug 3 at 22:26
Nicolas FRANCOIS
3,1121414
answered Aug 3 at 22:26
Nicolas FRANCOIS
3,1121414
answered Aug 3 at 22:26
Nicolas FRANCOIS
3,1121414
3,1121414
add a comment |Â
add a comment |Â
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Its not hard to prove it with induction.
– Calvin Khor
Aug 3 at 22:18
What you wrote looks good enough if only it was typeset in MathJax.
– Arnaud Mortier
Aug 3 at 22:20
Besides proving $fracd^kdx^k(x^k) = k!$ by induction on $k$, you could also formalize your argument directly: prove by induction on $n$ that whenever $1 le n le k$, then $fracd^ndx^n(x^k) = frack!(k-n)! x^k-n$.
– Daniel Schepler
Aug 3 at 22:25
A tutorial for typing mathematics in MathJax can be found here, by the way: math.meta.stackexchange.com/questions/5020/…
– Robert Howard
Aug 3 at 22:25