Simple question about a formula for sumsinequality of sumsFormula of Squaring Sums / IntegralsSolve easy sums with Binomial CoefficientWhy is this nested sum formula trueUsing a visual “proof” to show that $sum_n=1^infty left(frac 34 right)^n =1$Does an induction argument work when $n rightarrow infty$?Difficult rearrangement of summation terms to find analytical expressions for cumulant functionsquestion about sequences and seriesCalculate powers of sumsSimple question about Riemann zeta function
Is it legal for private citizens to "impound" e-scooters?
Why are off grid solar setups only 12, 24, 48 VDC?
Is a fighting a fallen friend with the help of a redeemed villain story too much for one book
kids pooling money for Lego League and taxes
Trapped in an ocean Temple in Minecraft?
What is the lowest-speed bogey a jet fighter can intercept/escort?
How could a thief buying plane tickets with stolen credit card details benefit personally?
Convert every file from JPEG to GIF in terminal
Why did Saturn V not head straight to the moon?
At what rate does the volume (velocity) of a note decay?
Terence Tao–type books in other fields?
Spoken encryption
Why was Sauron preparing for war instead of trying to find the ring?
Assuring luggage isn't lost with short layover
Commercial jet accompanied by small plane near Seattle
Marrying a second woman behind your wife's back: is it wrong and can Quran/Hadith prove this?
401(k) investment after being fired. Do I own it?
How to handle a player that cannot be convinced his actions are a problem for both GM and party
Problem in styling a monochrome plot
Writing a clean implementation of Rock, Paper, Scissors game in c++
Why can't my huge trees be chopped down?
Can I make a matrix from just a parts of the cells?
How to change the font style (not the size but the style) of algorithimc package
Heisenberg uncertainty principle in daily life
Simple question about a formula for sums
inequality of sumsFormula of Squaring Sums / IntegralsSolve easy sums with Binomial CoefficientWhy is this nested sum formula trueUsing a visual “proof” to show that $sum_n=1^infty left(frac 34 right)^n =1$Does an induction argument work when $n rightarrow infty$?Difficult rearrangement of summation terms to find analytical expressions for cumulant functionsquestion about sequences and seriesCalculate powers of sumsSimple question about Riemann zeta function
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Why is
$$
left(sum_i=1^na_iright)^2= sum_i=1^nsum_j=1^na_ia_j
$$
I guess it is fairly clear when you do it for $n=2$ for example, but I can't really proof it for all $n$ with induction.
calculus analysis summation
edited 9 hours ago
saulspatz
22.2k4 gold badges16 silver badges38 bronze badges
asked 9 hours ago
KingDingelingKingDingeling
45011 bronze badges
$endgroup$
add a comment |
$begingroup$
Why is
$$
left(sum_i=1^na_iright)^2= sum_i=1^nsum_j=1^na_ia_j
$$
I guess it is fairly clear when you do it for $n=2$ for example, but I can't really proof it for all $n$ with induction.
calculus analysis summation
edited 9 hours ago
saulspatz
22.2k4 gold badges16 silver badges38 bronze badges
asked 9 hours ago
KingDingelingKingDingeling
45011 bronze badges
$endgroup$
add a comment |
$begingroup$
Why is
$$
left(sum_i=1^na_iright)^2= sum_i=1^nsum_j=1^na_ia_j
$$
I guess it is fairly clear when you do it for $n=2$ for example, but I can't really proof it for all $n$ with induction.
calculus analysis summation
edited 9 hours ago
saulspatz
22.2k4 gold badges16 silver badges38 bronze badges
asked 9 hours ago
KingDingelingKingDingeling
45011 bronze badges
$endgroup$
Why is
$$
left(sum_i=1^na_iright)^2= sum_i=1^nsum_j=1^na_ia_j
$$
I guess it is fairly clear when you do it for $n=2$ for example, but I can't really proof it for all $n$ with induction.
calculus analysis summation
calculus analysis summation
edited 9 hours ago
saulspatz
22.2k4 gold badges16 silver badges38 bronze badges
asked 9 hours ago
KingDingelingKingDingeling
45011 bronze badges
edited 9 hours ago
saulspatz
22.2k4 gold badges16 silver badges38 bronze badges
asked 9 hours ago
KingDingelingKingDingeling
45011 bronze badges
edited 9 hours ago
saulspatz
22.2k4 gold badges16 silver badges38 bronze badges
edited 9 hours ago
saulspatz
22.2k4 gold badges16 silver badges38 bronze badges
edited 9 hours ago
saulspatz
22.2k4 gold badges16 silver badges38 bronze badges
22.2k4 gold badges16 silver badges38 bronze badges
asked 9 hours ago
KingDingelingKingDingeling
45011 bronze badges
asked 9 hours ago
KingDingelingKingDingeling
45011 bronze badges
asked 9 hours ago
KingDingelingKingDingeling
45011 bronze badges
45011 bronze badges
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
NB: I give a proof by induction in the following. It may be a good exercise in order to practice induction, but the best way to understand the formula really is Saulspatz' point of view.
We may prove this identity by induction on $n$. We will make use of the base case $n=2$, which you may check by yourself.
The heredity is given by the following. Let $n>2$ be given and assume that the property holds for $n-1$. Let $a_1,ldots, a_n$ be given real numbers, complex numbers, or elements of any commutative ring. We define $A_1 = a_1+ldots +a_n-1$ and $A_2 = a_n$. Using the base case $n=2$, we may write
$$left(sum_i=1^na_iright)^2=(A_1+A_2)^2=A_1^2 + A_1A_2+A_2A_1+A_2^2$$
Now, the induction hypothesis gives $A_1^2=sum_i=1^n-1sum_j=1^n-1a_ia_j$, and $A_2$ is no other than $a_n^2$.
As for $A_2A_1$, it is $sum_j=1^n-1a_na_j$. We may see it as the term "$i=n$" in order to regroup it inside the sum defining $A_1^2$. At this point, we have
$$left(sum_i=1^na_iright)^2= sum_i=1^nsum_j=1^n-1a_ia_j+A_1A_2+a_n^2$$
As you may know, because we are considering finite sums, we may invert the order of summations. Thus, this is also
$$left(sum_i=1^na_iright)^2= sum_j=1^n-1sum_i=1^na_ia_j+A_1A_2+a_n^2$$
But $A_1A_2$ is $sum_i=1^n-1a_ia_n=left(sum_i=1^na_ia_nright)-a_n^2$. Again, we identify the term "$j=n$" so that we may eventually regroup all the terms
$$left(sum_i=1^na_iright)^2= sum_i=1^nsum_j=1^na_ia_j$$
which is our desired identity.
answered 8 hours ago
SuzetSuzet
2,8326 silver badges27 bronze badges
$endgroup$
add a comment |
$begingroup$
On the left-hand side we have $$(a_1+a_2+cdots+a_n)(a_1+a_2+cdots+a_n)$$ How do we multiply this out? Pick a term in the first factor and a term in the second factor and multiply them. Add up the products for all possible choices of two terms. That's what the right-hand side says.
answered 9 hours ago
saulspatzsaulspatz
22.2k4 gold badges16 silver badges38 bronze badges
$endgroup$
add a comment |
Your Answer
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%2f3305875%2fsimple-question-about-a-formula-for-sums%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
NB: I give a proof by induction in the following. It may be a good exercise in order to practice induction, but the best way to understand the formula really is Saulspatz' point of view.
We may prove this identity by induction on $n$. We will make use of the base case $n=2$, which you may check by yourself.
The heredity is given by the following. Let $n>2$ be given and assume that the property holds for $n-1$. Let $a_1,ldots, a_n$ be given real numbers, complex numbers, or elements of any commutative ring. We define $A_1 = a_1+ldots +a_n-1$ and $A_2 = a_n$. Using the base case $n=2$, we may write
$$left(sum_i=1^na_iright)^2=(A_1+A_2)^2=A_1^2 + A_1A_2+A_2A_1+A_2^2$$
Now, the induction hypothesis gives $A_1^2=sum_i=1^n-1sum_j=1^n-1a_ia_j$, and $A_2$ is no other than $a_n^2$.
As for $A_2A_1$, it is $sum_j=1^n-1a_na_j$. We may see it as the term "$i=n$" in order to regroup it inside the sum defining $A_1^2$. At this point, we have
$$left(sum_i=1^na_iright)^2= sum_i=1^nsum_j=1^n-1a_ia_j+A_1A_2+a_n^2$$
As you may know, because we are considering finite sums, we may invert the order of summations. Thus, this is also
$$left(sum_i=1^na_iright)^2= sum_j=1^n-1sum_i=1^na_ia_j+A_1A_2+a_n^2$$
But $A_1A_2$ is $sum_i=1^n-1a_ia_n=left(sum_i=1^na_ia_nright)-a_n^2$. Again, we identify the term "$j=n$" so that we may eventually regroup all the terms
$$left(sum_i=1^na_iright)^2= sum_i=1^nsum_j=1^na_ia_j$$
which is our desired identity.
answered 8 hours ago
SuzetSuzet
2,8326 silver badges27 bronze badges
$endgroup$
add a comment |
$begingroup$
NB: I give a proof by induction in the following. It may be a good exercise in order to practice induction, but the best way to understand the formula really is Saulspatz' point of view.
We may prove this identity by induction on $n$. We will make use of the base case $n=2$, which you may check by yourself.
The heredity is given by the following. Let $n>2$ be given and assume that the property holds for $n-1$. Let $a_1,ldots, a_n$ be given real numbers, complex numbers, or elements of any commutative ring. We define $A_1 = a_1+ldots +a_n-1$ and $A_2 = a_n$. Using the base case $n=2$, we may write
$$left(sum_i=1^na_iright)^2=(A_1+A_2)^2=A_1^2 + A_1A_2+A_2A_1+A_2^2$$
Now, the induction hypothesis gives $A_1^2=sum_i=1^n-1sum_j=1^n-1a_ia_j$, and $A_2$ is no other than $a_n^2$.
As for $A_2A_1$, it is $sum_j=1^n-1a_na_j$. We may see it as the term "$i=n$" in order to regroup it inside the sum defining $A_1^2$. At this point, we have
$$left(sum_i=1^na_iright)^2= sum_i=1^nsum_j=1^n-1a_ia_j+A_1A_2+a_n^2$$
As you may know, because we are considering finite sums, we may invert the order of summations. Thus, this is also
$$left(sum_i=1^na_iright)^2= sum_j=1^n-1sum_i=1^na_ia_j+A_1A_2+a_n^2$$
But $A_1A_2$ is $sum_i=1^n-1a_ia_n=left(sum_i=1^na_ia_nright)-a_n^2$. Again, we identify the term "$j=n$" so that we may eventually regroup all the terms
$$left(sum_i=1^na_iright)^2= sum_i=1^nsum_j=1^na_ia_j$$
which is our desired identity.
answered 8 hours ago
SuzetSuzet
2,8326 silver badges27 bronze badges
$endgroup$
add a comment |
$begingroup$
NB: I give a proof by induction in the following. It may be a good exercise in order to practice induction, but the best way to understand the formula really is Saulspatz' point of view.
We may prove this identity by induction on $n$. We will make use of the base case $n=2$, which you may check by yourself.
The heredity is given by the following. Let $n>2$ be given and assume that the property holds for $n-1$. Let $a_1,ldots, a_n$ be given real numbers, complex numbers, or elements of any commutative ring. We define $A_1 = a_1+ldots +a_n-1$ and $A_2 = a_n$. Using the base case $n=2$, we may write
$$left(sum_i=1^na_iright)^2=(A_1+A_2)^2=A_1^2 + A_1A_2+A_2A_1+A_2^2$$
Now, the induction hypothesis gives $A_1^2=sum_i=1^n-1sum_j=1^n-1a_ia_j$, and $A_2$ is no other than $a_n^2$.
As for $A_2A_1$, it is $sum_j=1^n-1a_na_j$. We may see it as the term "$i=n$" in order to regroup it inside the sum defining $A_1^2$. At this point, we have
$$left(sum_i=1^na_iright)^2= sum_i=1^nsum_j=1^n-1a_ia_j+A_1A_2+a_n^2$$
As you may know, because we are considering finite sums, we may invert the order of summations. Thus, this is also
$$left(sum_i=1^na_iright)^2= sum_j=1^n-1sum_i=1^na_ia_j+A_1A_2+a_n^2$$
But $A_1A_2$ is $sum_i=1^n-1a_ia_n=left(sum_i=1^na_ia_nright)-a_n^2$. Again, we identify the term "$j=n$" so that we may eventually regroup all the terms
$$left(sum_i=1^na_iright)^2= sum_i=1^nsum_j=1^na_ia_j$$
which is our desired identity.
answered 8 hours ago
SuzetSuzet
2,8326 silver badges27 bronze badges
$endgroup$
NB: I give a proof by induction in the following. It may be a good exercise in order to practice induction, but the best way to understand the formula really is Saulspatz' point of view.
We may prove this identity by induction on $n$. We will make use of the base case $n=2$, which you may check by yourself.
The heredity is given by the following. Let $n>2$ be given and assume that the property holds for $n-1$. Let $a_1,ldots, a_n$ be given real numbers, complex numbers, or elements of any commutative ring. We define $A_1 = a_1+ldots +a_n-1$ and $A_2 = a_n$. Using the base case $n=2$, we may write
$$left(sum_i=1^na_iright)^2=(A_1+A_2)^2=A_1^2 + A_1A_2+A_2A_1+A_2^2$$
Now, the induction hypothesis gives $A_1^2=sum_i=1^n-1sum_j=1^n-1a_ia_j$, and $A_2$ is no other than $a_n^2$.
As for $A_2A_1$, it is $sum_j=1^n-1a_na_j$. We may see it as the term "$i=n$" in order to regroup it inside the sum defining $A_1^2$. At this point, we have
$$left(sum_i=1^na_iright)^2= sum_i=1^nsum_j=1^n-1a_ia_j+A_1A_2+a_n^2$$
As you may know, because we are considering finite sums, we may invert the order of summations. Thus, this is also
$$left(sum_i=1^na_iright)^2= sum_j=1^n-1sum_i=1^na_ia_j+A_1A_2+a_n^2$$
But $A_1A_2$ is $sum_i=1^n-1a_ia_n=left(sum_i=1^na_ia_nright)-a_n^2$. Again, we identify the term "$j=n$" so that we may eventually regroup all the terms
$$left(sum_i=1^na_iright)^2= sum_i=1^nsum_j=1^na_ia_j$$
which is our desired identity.
answered 8 hours ago
SuzetSuzet
2,8326 silver badges27 bronze badges
answered 8 hours ago
SuzetSuzet
2,8326 silver badges27 bronze badges
answered 8 hours ago
SuzetSuzet
2,8326 silver badges27 bronze badges
answered 8 hours ago
SuzetSuzet
2,8326 silver badges27 bronze badges
2,8326 silver badges27 bronze badges
add a comment |
add a comment |
$begingroup$
On the left-hand side we have $$(a_1+a_2+cdots+a_n)(a_1+a_2+cdots+a_n)$$ How do we multiply this out? Pick a term in the first factor and a term in the second factor and multiply them. Add up the products for all possible choices of two terms. That's what the right-hand side says.
answered 9 hours ago
saulspatzsaulspatz
22.2k4 gold badges16 silver badges38 bronze badges
$endgroup$
add a comment |
$begingroup$
On the left-hand side we have $$(a_1+a_2+cdots+a_n)(a_1+a_2+cdots+a_n)$$ How do we multiply this out? Pick a term in the first factor and a term in the second factor and multiply them. Add up the products for all possible choices of two terms. That's what the right-hand side says.
answered 9 hours ago
saulspatzsaulspatz
22.2k4 gold badges16 silver badges38 bronze badges
$endgroup$
add a comment |
$begingroup$
On the left-hand side we have $$(a_1+a_2+cdots+a_n)(a_1+a_2+cdots+a_n)$$ How do we multiply this out? Pick a term in the first factor and a term in the second factor and multiply them. Add up the products for all possible choices of two terms. That's what the right-hand side says.
answered 9 hours ago
saulspatzsaulspatz
22.2k4 gold badges16 silver badges38 bronze badges
$endgroup$
On the left-hand side we have $$(a_1+a_2+cdots+a_n)(a_1+a_2+cdots+a_n)$$ How do we multiply this out? Pick a term in the first factor and a term in the second factor and multiply them. Add up the products for all possible choices of two terms. That's what the right-hand side says.
answered 9 hours ago
saulspatzsaulspatz
22.2k4 gold badges16 silver badges38 bronze badges
answered 9 hours ago
saulspatzsaulspatz
22.2k4 gold badges16 silver badges38 bronze badges
answered 9 hours ago
saulspatzsaulspatz
22.2k4 gold badges16 silver badges38 bronze badges
answered 9 hours ago
saulspatzsaulspatz
22.2k4 gold badges16 silver badges38 bronze badges
22.2k4 gold badges16 silver badges38 bronze badges
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%2f3305875%2fsimple-question-about-a-formula-for-sums%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
