When was the generalization easier to prove than the specific case? [duplicate]
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This question already has an answer here:
Generalizing a problem to make it easier
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I distinctly remember from my long-ago undergraduate math that there were some interesting cases where a solution (proof) was sought for some specific thing but it wasn't easy to find - and in a few of these cases a generalization was made to some wider problem and that turned out to be provable (maybe not easily, but understandably) and of course that in turn solved the specific case.
But I don't remember any such situation and would like to know of one or more.
(I don't think the whole issue of the power of complex numbers - e.g., in the Fundamental Theorem of Algebra or in complex analysis vs real analysis is what I'm thinking of. I'm sure there were more closely focused cases of this phenomenon that I once knew.)
soft-question big-list examples
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marked as duplicate by Wojowu, Harry Gindi, Timothy Chow, Chris Godsil, YCor Feb 14 at 17:29
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Generalizing a problem to make it easier
45 answers
I distinctly remember from my long-ago undergraduate math that there were some interesting cases where a solution (proof) was sought for some specific thing but it wasn't easy to find - and in a few of these cases a generalization was made to some wider problem and that turned out to be provable (maybe not easily, but understandably) and of course that in turn solved the specific case.
But I don't remember any such situation and would like to know of one or more.
(I don't think the whole issue of the power of complex numbers - e.g., in the Fundamental Theorem of Algebra or in complex analysis vs real analysis is what I'm thinking of. I'm sure there were more closely focused cases of this phenomenon that I once knew.)
soft-question big-list examples
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marked as duplicate by Wojowu, Harry Gindi, Timothy Chow, Chris Godsil, YCor Feb 14 at 17:29
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
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mathoverflow.net/questions/40005/…
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– Wojowu
Feb 14 at 15:41
2
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@HarryGindi, your duplicate comment is a duplicate of @Wojowu's duplicate comment.
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– LSpice
Feb 14 at 16:06
5
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@LSpice It gets added automatically when you vote to close with the reason being duplicate.
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– Harry Gindi
Feb 14 at 16:11
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Harry's correct. I have first voted to close on the basis of this being better fit on Math.SE, but later I have noticed this is a duplicate of a question on MO. This is why we've got the repeated comment.
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– Wojowu
Feb 14 at 16:35
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Ah, thanks for the pointers to the duplicate, I didn't find that.
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– davidbak
Feb 14 at 18:09
add a comment |
$begingroup$
This question already has an answer here:
Generalizing a problem to make it easier
45 answers
I distinctly remember from my long-ago undergraduate math that there were some interesting cases where a solution (proof) was sought for some specific thing but it wasn't easy to find - and in a few of these cases a generalization was made to some wider problem and that turned out to be provable (maybe not easily, but understandably) and of course that in turn solved the specific case.
But I don't remember any such situation and would like to know of one or more.
(I don't think the whole issue of the power of complex numbers - e.g., in the Fundamental Theorem of Algebra or in complex analysis vs real analysis is what I'm thinking of. I'm sure there were more closely focused cases of this phenomenon that I once knew.)
soft-question big-list examples
$endgroup$
This question already has an answer here:
Generalizing a problem to make it easier
45 answers
I distinctly remember from my long-ago undergraduate math that there were some interesting cases where a solution (proof) was sought for some specific thing but it wasn't easy to find - and in a few of these cases a generalization was made to some wider problem and that turned out to be provable (maybe not easily, but understandably) and of course that in turn solved the specific case.
But I don't remember any such situation and would like to know of one or more.
(I don't think the whole issue of the power of complex numbers - e.g., in the Fundamental Theorem of Algebra or in complex analysis vs real analysis is what I'm thinking of. I'm sure there were more closely focused cases of this phenomenon that I once knew.)
This question already has an answer here:
Generalizing a problem to make it easier
45 answers
soft-question big-list examples
soft-question big-list examples
edited Feb 14 at 15:40
community wiki
davidbak
marked as duplicate by Wojowu, Harry Gindi, Timothy Chow, Chris Godsil, YCor Feb 14 at 17:29
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Wojowu, Harry Gindi, Timothy Chow, Chris Godsil, YCor Feb 14 at 17:29
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
$begingroup$
mathoverflow.net/questions/40005/…
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– Wojowu
Feb 14 at 15:41
2
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@HarryGindi, your duplicate comment is a duplicate of @Wojowu's duplicate comment.
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– LSpice
Feb 14 at 16:06
5
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@LSpice It gets added automatically when you vote to close with the reason being duplicate.
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– Harry Gindi
Feb 14 at 16:11
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Harry's correct. I have first voted to close on the basis of this being better fit on Math.SE, but later I have noticed this is a duplicate of a question on MO. This is why we've got the repeated comment.
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– Wojowu
Feb 14 at 16:35
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Ah, thanks for the pointers to the duplicate, I didn't find that.
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– davidbak
Feb 14 at 18:09
add a comment |
2
$begingroup$
mathoverflow.net/questions/40005/…
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– Wojowu
Feb 14 at 15:41
2
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@HarryGindi, your duplicate comment is a duplicate of @Wojowu's duplicate comment.
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– LSpice
Feb 14 at 16:06
5
$begingroup$
@LSpice It gets added automatically when you vote to close with the reason being duplicate.
$endgroup$
– Harry Gindi
Feb 14 at 16:11
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Harry's correct. I have first voted to close on the basis of this being better fit on Math.SE, but later I have noticed this is a duplicate of a question on MO. This is why we've got the repeated comment.
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– Wojowu
Feb 14 at 16:35
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Ah, thanks for the pointers to the duplicate, I didn't find that.
$endgroup$
– davidbak
Feb 14 at 18:09
2
2
$begingroup$
mathoverflow.net/questions/40005/…
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– Wojowu
Feb 14 at 15:41
$begingroup$
mathoverflow.net/questions/40005/…
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– Wojowu
Feb 14 at 15:41
2
2
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@HarryGindi, your duplicate comment is a duplicate of @Wojowu's duplicate comment.
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– LSpice
Feb 14 at 16:06
$begingroup$
@HarryGindi, your duplicate comment is a duplicate of @Wojowu's duplicate comment.
$endgroup$
– LSpice
Feb 14 at 16:06
5
5
$begingroup$
@LSpice It gets added automatically when you vote to close with the reason being duplicate.
$endgroup$
– Harry Gindi
Feb 14 at 16:11
$begingroup$
@LSpice It gets added automatically when you vote to close with the reason being duplicate.
$endgroup$
– Harry Gindi
Feb 14 at 16:11
$begingroup$
Harry's correct. I have first voted to close on the basis of this being better fit on Math.SE, but later I have noticed this is a duplicate of a question on MO. This is why we've got the repeated comment.
$endgroup$
– Wojowu
Feb 14 at 16:35
$begingroup$
Harry's correct. I have first voted to close on the basis of this being better fit on Math.SE, but later I have noticed this is a duplicate of a question on MO. This is why we've got the repeated comment.
$endgroup$
– Wojowu
Feb 14 at 16:35
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Ah, thanks for the pointers to the duplicate, I didn't find that.
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– davidbak
Feb 14 at 18:09
$begingroup$
Ah, thanks for the pointers to the duplicate, I didn't find that.
$endgroup$
– davidbak
Feb 14 at 18:09
add a comment |
2 Answers
2
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oldest
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Here's a very common type of example. Suppose you are interested in a certain convergent series $sum_{n=0}^infty a_n$. You look instead at the more general series $sum_{n=0}^infty a_n z^n$, to which you can apply the machinery of calculus, differential equations etc.,
and then specialize your result to $z=1$.
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add a comment |
$begingroup$
- There are many examples where this happens with properties concerning integers, where a property for a given integer may not be easy to prove, but the proof for the $n=0$ or $1$ case is easy, and induction is easy as well.
Alain Connes gave a great example of this on the radio you could even explain to a non-mathematician (actually that was the point): if I give you a chocolate bar with $8times 5$ squares, and you're allowed to cut it by breaking any piece into two pieces along a straight line (not intersecting the squares), what is the way of doing it that requires the least amount of cutting ? For $8times 5$ it's not clear, but if you generalize to $ntimes m$ and use induction, then it becomes quite clear.
Perhaps a more mathematical example of this phenomenon would be quite appreciated: many divisbility questions fit into this, e.g. prove that $2^{2n}+2$ is divisible by $3$
- A different family of examples comes from applications of group theory or ring theory to number theory; for instance Euler's theorem : $a^{varphi(n)} = 1 mod n$ if $aland n = 1$. I have never seen Euler's original proof of this, but seeing how complicated non-group-theoretic proofs of Fermat's little theorem are, I can only imagine that proving this only by number theory must be hard; while Lagrange's theorem is really easy and deals with this immediately if you know Bezout's theorem.
Another example in this family could be the fact that $nmid varphi(a^n-1)$; or $n! mid displaystyleprod_{i=0}^{n-1}(q^n-q^i)$ whenever $q$ is a prime power (though for these, unlike with the first family of examples, it may not be clear how to go from the specific instance to the generalization)
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The first example is terrific, thank you!
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– davidbak
Feb 14 at 18:10
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here's a very common type of example. Suppose you are interested in a certain convergent series $sum_{n=0}^infty a_n$. You look instead at the more general series $sum_{n=0}^infty a_n z^n$, to which you can apply the machinery of calculus, differential equations etc.,
and then specialize your result to $z=1$.
$endgroup$
add a comment |
$begingroup$
Here's a very common type of example. Suppose you are interested in a certain convergent series $sum_{n=0}^infty a_n$. You look instead at the more general series $sum_{n=0}^infty a_n z^n$, to which you can apply the machinery of calculus, differential equations etc.,
and then specialize your result to $z=1$.
$endgroup$
add a comment |
$begingroup$
Here's a very common type of example. Suppose you are interested in a certain convergent series $sum_{n=0}^infty a_n$. You look instead at the more general series $sum_{n=0}^infty a_n z^n$, to which you can apply the machinery of calculus, differential equations etc.,
and then specialize your result to $z=1$.
$endgroup$
Here's a very common type of example. Suppose you are interested in a certain convergent series $sum_{n=0}^infty a_n$. You look instead at the more general series $sum_{n=0}^infty a_n z^n$, to which you can apply the machinery of calculus, differential equations etc.,
and then specialize your result to $z=1$.
answered Feb 14 at 15:23
community wiki
Robert Israel
add a comment |
add a comment |
$begingroup$
- There are many examples where this happens with properties concerning integers, where a property for a given integer may not be easy to prove, but the proof for the $n=0$ or $1$ case is easy, and induction is easy as well.
Alain Connes gave a great example of this on the radio you could even explain to a non-mathematician (actually that was the point): if I give you a chocolate bar with $8times 5$ squares, and you're allowed to cut it by breaking any piece into two pieces along a straight line (not intersecting the squares), what is the way of doing it that requires the least amount of cutting ? For $8times 5$ it's not clear, but if you generalize to $ntimes m$ and use induction, then it becomes quite clear.
Perhaps a more mathematical example of this phenomenon would be quite appreciated: many divisbility questions fit into this, e.g. prove that $2^{2n}+2$ is divisible by $3$
- A different family of examples comes from applications of group theory or ring theory to number theory; for instance Euler's theorem : $a^{varphi(n)} = 1 mod n$ if $aland n = 1$. I have never seen Euler's original proof of this, but seeing how complicated non-group-theoretic proofs of Fermat's little theorem are, I can only imagine that proving this only by number theory must be hard; while Lagrange's theorem is really easy and deals with this immediately if you know Bezout's theorem.
Another example in this family could be the fact that $nmid varphi(a^n-1)$; or $n! mid displaystyleprod_{i=0}^{n-1}(q^n-q^i)$ whenever $q$ is a prime power (though for these, unlike with the first family of examples, it may not be clear how to go from the specific instance to the generalization)
$endgroup$
$begingroup$
The first example is terrific, thank you!
$endgroup$
– davidbak
Feb 14 at 18:10
add a comment |
$begingroup$
- There are many examples where this happens with properties concerning integers, where a property for a given integer may not be easy to prove, but the proof for the $n=0$ or $1$ case is easy, and induction is easy as well.
Alain Connes gave a great example of this on the radio you could even explain to a non-mathematician (actually that was the point): if I give you a chocolate bar with $8times 5$ squares, and you're allowed to cut it by breaking any piece into two pieces along a straight line (not intersecting the squares), what is the way of doing it that requires the least amount of cutting ? For $8times 5$ it's not clear, but if you generalize to $ntimes m$ and use induction, then it becomes quite clear.
Perhaps a more mathematical example of this phenomenon would be quite appreciated: many divisbility questions fit into this, e.g. prove that $2^{2n}+2$ is divisible by $3$
- A different family of examples comes from applications of group theory or ring theory to number theory; for instance Euler's theorem : $a^{varphi(n)} = 1 mod n$ if $aland n = 1$. I have never seen Euler's original proof of this, but seeing how complicated non-group-theoretic proofs of Fermat's little theorem are, I can only imagine that proving this only by number theory must be hard; while Lagrange's theorem is really easy and deals with this immediately if you know Bezout's theorem.
Another example in this family could be the fact that $nmid varphi(a^n-1)$; or $n! mid displaystyleprod_{i=0}^{n-1}(q^n-q^i)$ whenever $q$ is a prime power (though for these, unlike with the first family of examples, it may not be clear how to go from the specific instance to the generalization)
$endgroup$
$begingroup$
The first example is terrific, thank you!
$endgroup$
– davidbak
Feb 14 at 18:10
add a comment |
$begingroup$
- There are many examples where this happens with properties concerning integers, where a property for a given integer may not be easy to prove, but the proof for the $n=0$ or $1$ case is easy, and induction is easy as well.
Alain Connes gave a great example of this on the radio you could even explain to a non-mathematician (actually that was the point): if I give you a chocolate bar with $8times 5$ squares, and you're allowed to cut it by breaking any piece into two pieces along a straight line (not intersecting the squares), what is the way of doing it that requires the least amount of cutting ? For $8times 5$ it's not clear, but if you generalize to $ntimes m$ and use induction, then it becomes quite clear.
Perhaps a more mathematical example of this phenomenon would be quite appreciated: many divisbility questions fit into this, e.g. prove that $2^{2n}+2$ is divisible by $3$
- A different family of examples comes from applications of group theory or ring theory to number theory; for instance Euler's theorem : $a^{varphi(n)} = 1 mod n$ if $aland n = 1$. I have never seen Euler's original proof of this, but seeing how complicated non-group-theoretic proofs of Fermat's little theorem are, I can only imagine that proving this only by number theory must be hard; while Lagrange's theorem is really easy and deals with this immediately if you know Bezout's theorem.
Another example in this family could be the fact that $nmid varphi(a^n-1)$; or $n! mid displaystyleprod_{i=0}^{n-1}(q^n-q^i)$ whenever $q$ is a prime power (though for these, unlike with the first family of examples, it may not be clear how to go from the specific instance to the generalization)
$endgroup$
- There are many examples where this happens with properties concerning integers, where a property for a given integer may not be easy to prove, but the proof for the $n=0$ or $1$ case is easy, and induction is easy as well.
Alain Connes gave a great example of this on the radio you could even explain to a non-mathematician (actually that was the point): if I give you a chocolate bar with $8times 5$ squares, and you're allowed to cut it by breaking any piece into two pieces along a straight line (not intersecting the squares), what is the way of doing it that requires the least amount of cutting ? For $8times 5$ it's not clear, but if you generalize to $ntimes m$ and use induction, then it becomes quite clear.
Perhaps a more mathematical example of this phenomenon would be quite appreciated: many divisbility questions fit into this, e.g. prove that $2^{2n}+2$ is divisible by $3$
- A different family of examples comes from applications of group theory or ring theory to number theory; for instance Euler's theorem : $a^{varphi(n)} = 1 mod n$ if $aland n = 1$. I have never seen Euler's original proof of this, but seeing how complicated non-group-theoretic proofs of Fermat's little theorem are, I can only imagine that proving this only by number theory must be hard; while Lagrange's theorem is really easy and deals with this immediately if you know Bezout's theorem.
Another example in this family could be the fact that $nmid varphi(a^n-1)$; or $n! mid displaystyleprod_{i=0}^{n-1}(q^n-q^i)$ whenever $q$ is a prime power (though for these, unlike with the first family of examples, it may not be clear how to go from the specific instance to the generalization)
answered Feb 14 at 17:24
community wiki
Max
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The first example is terrific, thank you!
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– davidbak
Feb 14 at 18:10
add a comment |
$begingroup$
The first example is terrific, thank you!
$endgroup$
– davidbak
Feb 14 at 18:10
$begingroup$
The first example is terrific, thank you!
$endgroup$
– davidbak
Feb 14 at 18:10
$begingroup$
The first example is terrific, thank you!
$endgroup$
– davidbak
Feb 14 at 18:10
add a comment |
2
$begingroup$
mathoverflow.net/questions/40005/…
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– Wojowu
Feb 14 at 15:41
2
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@HarryGindi, your duplicate comment is a duplicate of @Wojowu's duplicate comment.
$endgroup$
– LSpice
Feb 14 at 16:06
5
$begingroup$
@LSpice It gets added automatically when you vote to close with the reason being duplicate.
$endgroup$
– Harry Gindi
Feb 14 at 16:11
$begingroup$
Harry's correct. I have first voted to close on the basis of this being better fit on Math.SE, but later I have noticed this is a duplicate of a question on MO. This is why we've got the repeated comment.
$endgroup$
– Wojowu
Feb 14 at 16:35
$begingroup$
Ah, thanks for the pointers to the duplicate, I didn't find that.
$endgroup$
– davidbak
Feb 14 at 18:09