Mixing
Limit of the $r$-norm $(sum_{i=0}^n (y_1,x_i)-(y_2,x_i))^r )^{1/r}$ as $r to infty$ is $ max_i...
$begingroup$
In a book I was reading about data mining they wrote
$$
lim_{r to infty} left(sum_{i=0}^n (y_1,x_i)-(y_2,x_i))^r right)^{1/r}
$$
is equal to
$$
max_{ i=1,2,...,n} left|(y_1,x_i)-(y_2,x_i)right|
$$
This is of course logical since the biggest outcome will dominate the equation but I wonder how did they prove it?
Thank you a lot
limits norm
edited Jan 6 at 18:15
Lord_Farin
15.5k636108
asked Oct 22 '18 at 7:57
Basil BassamBasil Bassam
337
$endgroup$
|
show 1 more comment
$begingroup$
In a book I was reading about data mining they wrote
$$
lim_{r to infty} left(sum_{i=0}^n (y_1,x_i)-(y_2,x_i))^r right)^{1/r}
$$
is equal to
$$
max_{ i=1,2,...,n} left|(y_1,x_i)-(y_2,x_i)right|
$$
This is of course logical since the biggest outcome will dominate the equation but I wonder how did they prove it?
Thank you a lot
limits norm
edited Jan 6 at 18:15
Lord_Farin
15.5k636108
asked Oct 22 '18 at 7:57
Basil BassamBasil Bassam
337
$endgroup$
$begingroup$
My laptop is not working and that's the most thing I can do with my mobile :( sorry, can you edit it if possible
$endgroup$
– Basil Bassam
Oct 22 '18 at 8:01
1
$begingroup$
With your formatting, it's a bit tricky to figure out the correct equation, see if I guessed correctly
$endgroup$
– Matti P.
Oct 22 '18 at 8:07
$begingroup$
It is correct thank you very much
$endgroup$
– Basil Bassam
Oct 22 '18 at 8:10
$begingroup$
Any ideas how to prove this?
$endgroup$
– Basil Bassam
Oct 22 '18 at 8:15
$begingroup$
By the way, should the sum be inside the ()^(1/r) ? That way it would make more sense ...
$endgroup$
– Matti P.
Oct 22 '18 at 8:20
|
show 1 more comment
$begingroup$
In a book I was reading about data mining they wrote
$$
lim_{r to infty} left(sum_{i=0}^n (y_1,x_i)-(y_2,x_i))^r right)^{1/r}
$$
is equal to
$$
max_{ i=1,2,...,n} left|(y_1,x_i)-(y_2,x_i)right|
$$
This is of course logical since the biggest outcome will dominate the equation but I wonder how did they prove it?
Thank you a lot
limits norm
edited Jan 6 at 18:15
Lord_Farin
15.5k636108
asked Oct 22 '18 at 7:57
Basil BassamBasil Bassam
337
$endgroup$
In a book I was reading about data mining they wrote
$$
lim_{r to infty} left(sum_{i=0}^n (y_1,x_i)-(y_2,x_i))^r right)^{1/r}
$$
is equal to
$$
max_{ i=1,2,...,n} left|(y_1,x_i)-(y_2,x_i)right|
$$
This is of course logical since the biggest outcome will dominate the equation but I wonder how did they prove it?
Thank you a lot
limits norm
limits norm
edited Jan 6 at 18:15
Lord_Farin
15.5k636108
asked Oct 22 '18 at 7:57
Basil BassamBasil Bassam
337
edited Jan 6 at 18:15
Lord_Farin
15.5k636108
asked Oct 22 '18 at 7:57
Basil BassamBasil Bassam
337
edited Jan 6 at 18:15
Lord_Farin
15.5k636108
edited Jan 6 at 18:15
Lord_Farin
15.5k636108
edited Jan 6 at 18:15
Lord_Farin
15.5k636108
15.5k636108
asked Oct 22 '18 at 7:57
Basil BassamBasil Bassam
337
asked Oct 22 '18 at 7:57
Basil BassamBasil Bassam
337
asked Oct 22 '18 at 7:57
Basil BassamBasil Bassam
337
337
$begingroup$
My laptop is not working and that's the most thing I can do with my mobile :( sorry, can you edit it if possible
$endgroup$
– Basil Bassam
Oct 22 '18 at 8:01
1
$begingroup$
With your formatting, it's a bit tricky to figure out the correct equation, see if I guessed correctly
$endgroup$
– Matti P.
Oct 22 '18 at 8:07
$begingroup$
It is correct thank you very much
$endgroup$
– Basil Bassam
Oct 22 '18 at 8:10
$begingroup$
Any ideas how to prove this?
$endgroup$
– Basil Bassam
Oct 22 '18 at 8:15
$begingroup$
By the way, should the sum be inside the ()^(1/r) ? That way it would make more sense ...
$endgroup$
– Matti P.
Oct 22 '18 at 8:20
|
show 1 more comment
$begingroup$
My laptop is not working and that's the most thing I can do with my mobile :( sorry, can you edit it if possible
$endgroup$
– Basil Bassam
Oct 22 '18 at 8:01
1
$begingroup$
With your formatting, it's a bit tricky to figure out the correct equation, see if I guessed correctly
$endgroup$
– Matti P.
Oct 22 '18 at 8:07
$begingroup$
It is correct thank you very much
$endgroup$
– Basil Bassam
Oct 22 '18 at 8:10
$begingroup$
Any ideas how to prove this?
$endgroup$
– Basil Bassam
Oct 22 '18 at 8:15
$begingroup$
By the way, should the sum be inside the ()^(1/r) ? That way it would make more sense ...
$endgroup$
– Matti P.
Oct 22 '18 at 8:20
$begingroup$
My laptop is not working and that's the most thing I can do with my mobile :( sorry, can you edit it if possible
$endgroup$
– Basil Bassam
Oct 22 '18 at 8:01
$begingroup$
My laptop is not working and that's the most thing I can do with my mobile :( sorry, can you edit it if possible
$endgroup$
– Basil Bassam
Oct 22 '18 at 8:01
1
1
$begingroup$
With your formatting, it's a bit tricky to figure out the correct equation, see if I guessed correctly
$endgroup$
– Matti P.
Oct 22 '18 at 8:07
$begingroup$
With your formatting, it's a bit tricky to figure out the correct equation, see if I guessed correctly
$endgroup$
– Matti P.
Oct 22 '18 at 8:07
$begingroup$
It is correct thank you very much
$endgroup$
– Basil Bassam
Oct 22 '18 at 8:10
$begingroup$
It is correct thank you very much
$endgroup$
– Basil Bassam
Oct 22 '18 at 8:10
$begingroup$
Any ideas how to prove this?
$endgroup$
– Basil Bassam
Oct 22 '18 at 8:15
$begingroup$
Any ideas how to prove this?
$endgroup$
– Basil Bassam
Oct 22 '18 at 8:15
$begingroup$
By the way, should the sum be inside the ()^(1/r) ? That way it would make more sense ...
$endgroup$
– Matti P.
Oct 22 '18 at 8:20
$begingroup$
By the way, should the sum be inside the ()^(1/r) ? That way it would make more sense ...
$endgroup$
– Matti P.
Oct 22 '18 at 8:20
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Essentially we are considering a list of numbers $a_i$ and the $L_{infty}$ norm:
$$
L_{infty} = lim_{r to infty} left[ sum_{i} a_i^r right]^{1/r}
$$
Some entries in $a_i$ are smaller and some are larger. When you put all the entries to a large power (for example $r=100$), the differences become even more pronounced. Values that are close to $1$ remain close to $1$, but larger values get pushed towards infinity. So when you sum these up, the sum is basically only the sum of the large numbers, because a very large number plus a very small number is approximately equal to the large number. And remember that pushing the exponent up makes these differences even more pronounced.
Therefore, it's relatively intuitive that the value in the end turns out to be just the maximum absolute value in the list
$$
L_{infty} = lim_{r to infty} left[ sum_{i} a_i^r right]^{1/r} = max_i left| a_i right|
$$
As an example, let's take
$$
a = [0.2, 1.5, 2, 7, 25] qquad text{and} qquad r = 10
$$
then
$$
a^r approx [1.024 times 10^{-7} ,quad 57.6, quad 1024 282times 10^6,quad 95times 10^{12}]
$$
The first entries pale in comparison to the last one, and the sum is
$$
sum a_i^r approx 9.5368 times 10^{13} Rightarrow sqrt[10]{sum a_i^r} approx 25
$$
which is the maximum of the values.
This was not a rigorous proof, but surely you'll understand the intuition.
answered Oct 22 '18 at 8:33
Matti P.Matti P.
1,881413
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2965692%2flimit-of-the-r-norm-sum-i-0n-y-1-x-i-y-2-x-ir-1-r-as-r-to%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Essentially we are considering a list of numbers $a_i$ and the $L_{infty}$ norm:
$$
L_{infty} = lim_{r to infty} left[ sum_{i} a_i^r right]^{1/r}
$$
Some entries in $a_i$ are smaller and some are larger. When you put all the entries to a large power (for example $r=100$), the differences become even more pronounced. Values that are close to $1$ remain close to $1$, but larger values get pushed towards infinity. So when you sum these up, the sum is basically only the sum of the large numbers, because a very large number plus a very small number is approximately equal to the large number. And remember that pushing the exponent up makes these differences even more pronounced.
Therefore, it's relatively intuitive that the value in the end turns out to be just the maximum absolute value in the list
$$
L_{infty} = lim_{r to infty} left[ sum_{i} a_i^r right]^{1/r} = max_i left| a_i right|
$$
As an example, let's take
$$
a = [0.2, 1.5, 2, 7, 25] qquad text{and} qquad r = 10
$$
then
$$
a^r approx [1.024 times 10^{-7} ,quad 57.6, quad 1024 282times 10^6,quad 95times 10^{12}]
$$
The first entries pale in comparison to the last one, and the sum is
$$
sum a_i^r approx 9.5368 times 10^{13} Rightarrow sqrt[10]{sum a_i^r} approx 25
$$
which is the maximum of the values.
This was not a rigorous proof, but surely you'll understand the intuition.
answered Oct 22 '18 at 8:33
Matti P.Matti P.
1,881413
$endgroup$
add a comment |
$begingroup$
Essentially we are considering a list of numbers $a_i$ and the $L_{infty}$ norm:
$$
L_{infty} = lim_{r to infty} left[ sum_{i} a_i^r right]^{1/r}
$$
Some entries in $a_i$ are smaller and some are larger. When you put all the entries to a large power (for example $r=100$), the differences become even more pronounced. Values that are close to $1$ remain close to $1$, but larger values get pushed towards infinity. So when you sum these up, the sum is basically only the sum of the large numbers, because a very large number plus a very small number is approximately equal to the large number. And remember that pushing the exponent up makes these differences even more pronounced.
Therefore, it's relatively intuitive that the value in the end turns out to be just the maximum absolute value in the list
$$
L_{infty} = lim_{r to infty} left[ sum_{i} a_i^r right]^{1/r} = max_i left| a_i right|
$$
As an example, let's take
$$
a = [0.2, 1.5, 2, 7, 25] qquad text{and} qquad r = 10
$$
then
$$
a^r approx [1.024 times 10^{-7} ,quad 57.6, quad 1024 282times 10^6,quad 95times 10^{12}]
$$
The first entries pale in comparison to the last one, and the sum is
$$
sum a_i^r approx 9.5368 times 10^{13} Rightarrow sqrt[10]{sum a_i^r} approx 25
$$
which is the maximum of the values.
This was not a rigorous proof, but surely you'll understand the intuition.
answered Oct 22 '18 at 8:33
Matti P.Matti P.
1,881413
$endgroup$
add a comment |
$begingroup$
Essentially we are considering a list of numbers $a_i$ and the $L_{infty}$ norm:
$$
L_{infty} = lim_{r to infty} left[ sum_{i} a_i^r right]^{1/r}
$$
Some entries in $a_i$ are smaller and some are larger. When you put all the entries to a large power (for example $r=100$), the differences become even more pronounced. Values that are close to $1$ remain close to $1$, but larger values get pushed towards infinity. So when you sum these up, the sum is basically only the sum of the large numbers, because a very large number plus a very small number is approximately equal to the large number. And remember that pushing the exponent up makes these differences even more pronounced.
Therefore, it's relatively intuitive that the value in the end turns out to be just the maximum absolute value in the list
$$
L_{infty} = lim_{r to infty} left[ sum_{i} a_i^r right]^{1/r} = max_i left| a_i right|
$$
As an example, let's take
$$
a = [0.2, 1.5, 2, 7, 25] qquad text{and} qquad r = 10
$$
then
$$
a^r approx [1.024 times 10^{-7} ,quad 57.6, quad 1024 282times 10^6,quad 95times 10^{12}]
$$
The first entries pale in comparison to the last one, and the sum is
$$
sum a_i^r approx 9.5368 times 10^{13} Rightarrow sqrt[10]{sum a_i^r} approx 25
$$
which is the maximum of the values.
This was not a rigorous proof, but surely you'll understand the intuition.
answered Oct 22 '18 at 8:33
Matti P.Matti P.
1,881413
$endgroup$
Essentially we are considering a list of numbers $a_i$ and the $L_{infty}$ norm:
$$
L_{infty} = lim_{r to infty} left[ sum_{i} a_i^r right]^{1/r}
$$
Some entries in $a_i$ are smaller and some are larger. When you put all the entries to a large power (for example $r=100$), the differences become even more pronounced. Values that are close to $1$ remain close to $1$, but larger values get pushed towards infinity. So when you sum these up, the sum is basically only the sum of the large numbers, because a very large number plus a very small number is approximately equal to the large number. And remember that pushing the exponent up makes these differences even more pronounced.
Therefore, it's relatively intuitive that the value in the end turns out to be just the maximum absolute value in the list
$$
L_{infty} = lim_{r to infty} left[ sum_{i} a_i^r right]^{1/r} = max_i left| a_i right|
$$
As an example, let's take
$$
a = [0.2, 1.5, 2, 7, 25] qquad text{and} qquad r = 10
$$
then
$$
a^r approx [1.024 times 10^{-7} ,quad 57.6, quad 1024 282times 10^6,quad 95times 10^{12}]
$$
The first entries pale in comparison to the last one, and the sum is
$$
sum a_i^r approx 9.5368 times 10^{13} Rightarrow sqrt[10]{sum a_i^r} approx 25
$$
which is the maximum of the values.
This was not a rigorous proof, but surely you'll understand the intuition.
answered Oct 22 '18 at 8:33
Matti P.Matti P.
1,881413
edited Oct 22 '18 at 8:43
answered Oct 22 '18 at 8:33
Matti P.Matti P.
1,881413
answered Oct 22 '18 at 8:33
Matti P.Matti P.
1,881413
answered Oct 22 '18 at 8:33
Matti P.Matti P.
1,881413
1,881413
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2965692%2flimit-of-the-r-norm-sum-i-0n-y-1-x-i-y-2-x-ir-1-r-as-r-to%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
My laptop is not working and that's the most thing I can do with my mobile :( sorry, can you edit it if possible
$endgroup$
– Basil Bassam
Oct 22 '18 at 8:01
1
$begingroup$
With your formatting, it's a bit tricky to figure out the correct equation, see if I guessed correctly
$endgroup$
– Matti P.
Oct 22 '18 at 8:07
$begingroup$
It is correct thank you very much
$endgroup$
– Basil Bassam
Oct 22 '18 at 8:10
$begingroup$
Any ideas how to prove this?
$endgroup$
– Basil Bassam
Oct 22 '18 at 8:15
$begingroup$
By the way, should the sum be inside the ()^(1/r) ? That way it would make more sense ...
$endgroup$
– Matti P.
Oct 22 '18 at 8:20