Mixing
A series question whose indices involves greatest integer function
$begingroup$
Let $x$ be a real number and $lfloor{x}rfloor$ denote the greatest integer less than or equal to $x$. Let $a$ and $n$ be nonnegative integers such that $ngeq a+1$. Prove that
$$ sum_{k=a}^{n} f(k) - sum_{k=a}^{n} (-1)^{k} f(k) = 2 sum_{k=lfloor{frac{a+2}{2}}rfloor}^{lfloor{frac{n+1}{2}}rfloor} f(2k-1).$$
I proved this by considering the situations where $a$ and $n$ takes even or odd values respectively. Are there any other way to show the equation is true? I generally have trouble with series whose indices involves greatest integer function, are there fast techniques to solve them?
sequences-and-series floor-function
asked Jan 28 at 10:14
Bright ChancellorBright Chancellor
241313
$endgroup$
add a comment |
$begingroup$
Let $x$ be a real number and $lfloor{x}rfloor$ denote the greatest integer less than or equal to $x$. Let $a$ and $n$ be nonnegative integers such that $ngeq a+1$. Prove that
$$ sum_{k=a}^{n} f(k) - sum_{k=a}^{n} (-1)^{k} f(k) = 2 sum_{k=lfloor{frac{a+2}{2}}rfloor}^{lfloor{frac{n+1}{2}}rfloor} f(2k-1).$$
I proved this by considering the situations where $a$ and $n$ takes even or odd values respectively. Are there any other way to show the equation is true? I generally have trouble with series whose indices involves greatest integer function, are there fast techniques to solve them?
sequences-and-series floor-function
asked Jan 28 at 10:14
Bright ChancellorBright Chancellor
241313
$endgroup$
add a comment |
$begingroup$
Let $x$ be a real number and $lfloor{x}rfloor$ denote the greatest integer less than or equal to $x$. Let $a$ and $n$ be nonnegative integers such that $ngeq a+1$. Prove that
$$ sum_{k=a}^{n} f(k) - sum_{k=a}^{n} (-1)^{k} f(k) = 2 sum_{k=lfloor{frac{a+2}{2}}rfloor}^{lfloor{frac{n+1}{2}}rfloor} f(2k-1).$$
I proved this by considering the situations where $a$ and $n$ takes even or odd values respectively. Are there any other way to show the equation is true? I generally have trouble with series whose indices involves greatest integer function, are there fast techniques to solve them?
sequences-and-series floor-function
asked Jan 28 at 10:14
Bright ChancellorBright Chancellor
241313
$endgroup$
Let $x$ be a real number and $lfloor{x}rfloor$ denote the greatest integer less than or equal to $x$. Let $a$ and $n$ be nonnegative integers such that $ngeq a+1$. Prove that
$$ sum_{k=a}^{n} f(k) - sum_{k=a}^{n} (-1)^{k} f(k) = 2 sum_{k=lfloor{frac{a+2}{2}}rfloor}^{lfloor{frac{n+1}{2}}rfloor} f(2k-1).$$
I proved this by considering the situations where $a$ and $n$ takes even or odd values respectively. Are there any other way to show the equation is true? I generally have trouble with series whose indices involves greatest integer function, are there fast techniques to solve them?
sequences-and-series floor-function
sequences-and-series floor-function
asked Jan 28 at 10:14
Bright ChancellorBright Chancellor
241313
asked Jan 28 at 10:14
Bright ChancellorBright Chancellor
241313
asked Jan 28 at 10:14
Bright ChancellorBright Chancellor
241313
asked Jan 28 at 10:14
Bright ChancellorBright Chancellor
241313
asked Jan 28 at 10:14
Bright ChancellorBright Chancellor
241313
241313
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The index of summation is ranging as
$$
a le k le n
$$
Replace it with its even and odd components
$$
k = 2j - iquad left| {,i = 0,1} right.
$$
Then for the even component it shall be
$$
eqalign{
& i = 0quad Rightarrow quad a le 2j le nquad Rightarrow quad leftlceil {{a over 2}} rightrceil le j le leftlfloor {{n over 2}} rightrfloor
quad Rightarrow cr
& Rightarrow quad leftlfloor {{{a + 1} over 2}} rightrfloor le j le leftlfloor {{n over 2}} rightrfloor cr}
$$
and for the odd
$$
eqalign{
& i = 1quad Rightarrow quad a le 2j - 1 le nquad Rightarrow quad leftlceil {{{a + 1} over 2}} rightrceil le j le
leftlfloor {{{n + 1} over 2}} rightrfloor quad Rightarrow cr
& Rightarrow quad leftlfloor {{{a + 2} over 2}} rightrfloor le j le leftlfloor {{{n + 1} over 2}} rightrfloor cr}
$$
Note that for the lower bound we shall employ the ceiling and the floor for the upper. Then we can convert the ceiling to floor.
The sum over $k$ of $f(k)$ can be split into the sum over the even and the odd component
$$
eqalign{
& sumlimits_{a, le ,k, le ;n} {f(k)} = sumlimits_{a, le ,k, le ;n} {left. {f(k),} right|_{,k = 2j} + left. {f(k),} right|_{,k = 2j - 1} } = cr
& = sumlimits_{a, le ,2j, le ;n} {f(2j)} + sumlimits_{a, le ,2j - 1, le ;n} {f(2j - 1)} = cr
& = sumlimits_{leftlfloor {{{a + 1} over 2}} rightrfloor le j le leftlfloor {{n over 2}} rightrfloor } {f(2j)}
+ sumlimits_{leftlfloor {{{a + 2} over 2}} rightrfloor le j le leftlfloor {{{n + 1} over 2}} rightrfloor } {f(2j - 1)} cr}
$$
and same for that of $(-1)^k f(k)$
$$
eqalign{
& sumlimits_{a, le ,k, le ;n} {left( { - 1} right)^{,k} f(k)}
= sumlimits_{leftlfloor {{{a + 1} over 2}} rightrfloor le j le leftlfloor {{n over 2}} rightrfloor } {left( { - 1} right)^{,2j} f(2j)}
+ sumlimits_{leftlfloor {{{a + 2} over 2}} rightrfloor le j le leftlfloor {{{n + 1} over 2}} rightrfloor } {left( { - 1} right)^{,2j - 1} f(2j - 1)} = cr
& = sumlimits_{leftlfloor {{{a + 1} over 2}} rightrfloor le j le leftlfloor {{n over 2}} rightrfloor } {f(2j)}
- sumlimits_{leftlfloor {{{a + 2} over 2}} rightrfloor le j le leftlfloor {{{n + 1} over 2}} rightrfloor } {f(2j - 1)} cr}
$$
Now, just subtract the two above to confirm the thesis.
--- addendum ----
Concerning your request on how to transform ceiling <-> floor, consider that
$$
eqalign{
& n = 2leftlceil {{n over 2}} rightrceil + 2left{ {{n over 2}} right} = 2q + iquad left| matrix{
;n,q in Z hfill cr
,i = 0,1 hfill cr} right. cr
& leftlceil {{n over 2}} rightrceil = leftlceil {{{2q + i} over 2}} rightrceil
= leftlceil {q + {i over 2}} rightrceil = q + leftlceil {{i over 2}} rightrceil = q + i cr
& leftlfloor {{{n + 1} over 2}} rightrfloor = leftlfloor {{{2q + 1 + i} over 2}} rightrfloor
= leftlfloor {q + {{1 + i} over 2}} rightrfloor = q + leftlfloor {{{1 + i} over 2}} rightrfloor = q + i cr}
$$
and, in general
$$
leftlceil {{n over m}} rightrceil = leftlfloor {{{n + m - 1} over m}} rightrfloor quad left| matrix{
;n,m in Z hfill cr
;1 le m hfill cr} right.
$$
re. for details to this article.
answered Jan 30 at 12:05
G CabG Cab
20.4k31341
$endgroup$
$begingroup$
How Did you convert the ceiling function to the floor? That is where I stuck actually.
$endgroup$
– Bright Chancellor
Jan 31 at 8:19
1
$begingroup$
@BrightChancellor: put an addendum to answer to this
$endgroup$
– G Cab
Jan 31 at 11:16
$begingroup$
I got it, thanks.
$endgroup$
– Bright Chancellor
Jan 31 at 11:22
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%2f3090685%2fa-series-question-whose-indices-involves-greatest-integer-function%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
$begingroup$
The index of summation is ranging as
$$
a le k le n
$$
Replace it with its even and odd components
$$
k = 2j - iquad left| {,i = 0,1} right.
$$
Then for the even component it shall be
$$
eqalign{
& i = 0quad Rightarrow quad a le 2j le nquad Rightarrow quad leftlceil {{a over 2}} rightrceil le j le leftlfloor {{n over 2}} rightrfloor
quad Rightarrow cr
& Rightarrow quad leftlfloor {{{a + 1} over 2}} rightrfloor le j le leftlfloor {{n over 2}} rightrfloor cr}
$$
and for the odd
$$
eqalign{
& i = 1quad Rightarrow quad a le 2j - 1 le nquad Rightarrow quad leftlceil {{{a + 1} over 2}} rightrceil le j le
leftlfloor {{{n + 1} over 2}} rightrfloor quad Rightarrow cr
& Rightarrow quad leftlfloor {{{a + 2} over 2}} rightrfloor le j le leftlfloor {{{n + 1} over 2}} rightrfloor cr}
$$
Note that for the lower bound we shall employ the ceiling and the floor for the upper. Then we can convert the ceiling to floor.
The sum over $k$ of $f(k)$ can be split into the sum over the even and the odd component
$$
eqalign{
& sumlimits_{a, le ,k, le ;n} {f(k)} = sumlimits_{a, le ,k, le ;n} {left. {f(k),} right|_{,k = 2j} + left. {f(k),} right|_{,k = 2j - 1} } = cr
& = sumlimits_{a, le ,2j, le ;n} {f(2j)} + sumlimits_{a, le ,2j - 1, le ;n} {f(2j - 1)} = cr
& = sumlimits_{leftlfloor {{{a + 1} over 2}} rightrfloor le j le leftlfloor {{n over 2}} rightrfloor } {f(2j)}
+ sumlimits_{leftlfloor {{{a + 2} over 2}} rightrfloor le j le leftlfloor {{{n + 1} over 2}} rightrfloor } {f(2j - 1)} cr}
$$
and same for that of $(-1)^k f(k)$
$$
eqalign{
& sumlimits_{a, le ,k, le ;n} {left( { - 1} right)^{,k} f(k)}
= sumlimits_{leftlfloor {{{a + 1} over 2}} rightrfloor le j le leftlfloor {{n over 2}} rightrfloor } {left( { - 1} right)^{,2j} f(2j)}
+ sumlimits_{leftlfloor {{{a + 2} over 2}} rightrfloor le j le leftlfloor {{{n + 1} over 2}} rightrfloor } {left( { - 1} right)^{,2j - 1} f(2j - 1)} = cr
& = sumlimits_{leftlfloor {{{a + 1} over 2}} rightrfloor le j le leftlfloor {{n over 2}} rightrfloor } {f(2j)}
- sumlimits_{leftlfloor {{{a + 2} over 2}} rightrfloor le j le leftlfloor {{{n + 1} over 2}} rightrfloor } {f(2j - 1)} cr}
$$
Now, just subtract the two above to confirm the thesis.
--- addendum ----
Concerning your request on how to transform ceiling <-> floor, consider that
$$
eqalign{
& n = 2leftlceil {{n over 2}} rightrceil + 2left{ {{n over 2}} right} = 2q + iquad left| matrix{
;n,q in Z hfill cr
,i = 0,1 hfill cr} right. cr
& leftlceil {{n over 2}} rightrceil = leftlceil {{{2q + i} over 2}} rightrceil
= leftlceil {q + {i over 2}} rightrceil = q + leftlceil {{i over 2}} rightrceil = q + i cr
& leftlfloor {{{n + 1} over 2}} rightrfloor = leftlfloor {{{2q + 1 + i} over 2}} rightrfloor
= leftlfloor {q + {{1 + i} over 2}} rightrfloor = q + leftlfloor {{{1 + i} over 2}} rightrfloor = q + i cr}
$$
and, in general
$$
leftlceil {{n over m}} rightrceil = leftlfloor {{{n + m - 1} over m}} rightrfloor quad left| matrix{
;n,m in Z hfill cr
;1 le m hfill cr} right.
$$
re. for details to this article.
answered Jan 30 at 12:05
G CabG Cab
20.4k31341
$endgroup$
$begingroup$
How Did you convert the ceiling function to the floor? That is where I stuck actually.
$endgroup$
– Bright Chancellor
Jan 31 at 8:19
1
$begingroup$
@BrightChancellor: put an addendum to answer to this
$endgroup$
– G Cab
Jan 31 at 11:16
$begingroup$
I got it, thanks.
$endgroup$
– Bright Chancellor
Jan 31 at 11:22
add a comment |
$begingroup$
The index of summation is ranging as
$$
a le k le n
$$
Replace it with its even and odd components
$$
k = 2j - iquad left| {,i = 0,1} right.
$$
Then for the even component it shall be
$$
eqalign{
& i = 0quad Rightarrow quad a le 2j le nquad Rightarrow quad leftlceil {{a over 2}} rightrceil le j le leftlfloor {{n over 2}} rightrfloor
quad Rightarrow cr
& Rightarrow quad leftlfloor {{{a + 1} over 2}} rightrfloor le j le leftlfloor {{n over 2}} rightrfloor cr}
$$
and for the odd
$$
eqalign{
& i = 1quad Rightarrow quad a le 2j - 1 le nquad Rightarrow quad leftlceil {{{a + 1} over 2}} rightrceil le j le
leftlfloor {{{n + 1} over 2}} rightrfloor quad Rightarrow cr
& Rightarrow quad leftlfloor {{{a + 2} over 2}} rightrfloor le j le leftlfloor {{{n + 1} over 2}} rightrfloor cr}
$$
Note that for the lower bound we shall employ the ceiling and the floor for the upper. Then we can convert the ceiling to floor.
The sum over $k$ of $f(k)$ can be split into the sum over the even and the odd component
$$
eqalign{
& sumlimits_{a, le ,k, le ;n} {f(k)} = sumlimits_{a, le ,k, le ;n} {left. {f(k),} right|_{,k = 2j} + left. {f(k),} right|_{,k = 2j - 1} } = cr
& = sumlimits_{a, le ,2j, le ;n} {f(2j)} + sumlimits_{a, le ,2j - 1, le ;n} {f(2j - 1)} = cr
& = sumlimits_{leftlfloor {{{a + 1} over 2}} rightrfloor le j le leftlfloor {{n over 2}} rightrfloor } {f(2j)}
+ sumlimits_{leftlfloor {{{a + 2} over 2}} rightrfloor le j le leftlfloor {{{n + 1} over 2}} rightrfloor } {f(2j - 1)} cr}
$$
and same for that of $(-1)^k f(k)$
$$
eqalign{
& sumlimits_{a, le ,k, le ;n} {left( { - 1} right)^{,k} f(k)}
= sumlimits_{leftlfloor {{{a + 1} over 2}} rightrfloor le j le leftlfloor {{n over 2}} rightrfloor } {left( { - 1} right)^{,2j} f(2j)}
+ sumlimits_{leftlfloor {{{a + 2} over 2}} rightrfloor le j le leftlfloor {{{n + 1} over 2}} rightrfloor } {left( { - 1} right)^{,2j - 1} f(2j - 1)} = cr
& = sumlimits_{leftlfloor {{{a + 1} over 2}} rightrfloor le j le leftlfloor {{n over 2}} rightrfloor } {f(2j)}
- sumlimits_{leftlfloor {{{a + 2} over 2}} rightrfloor le j le leftlfloor {{{n + 1} over 2}} rightrfloor } {f(2j - 1)} cr}
$$
Now, just subtract the two above to confirm the thesis.
--- addendum ----
Concerning your request on how to transform ceiling <-> floor, consider that
$$
eqalign{
& n = 2leftlceil {{n over 2}} rightrceil + 2left{ {{n over 2}} right} = 2q + iquad left| matrix{
;n,q in Z hfill cr
,i = 0,1 hfill cr} right. cr
& leftlceil {{n over 2}} rightrceil = leftlceil {{{2q + i} over 2}} rightrceil
= leftlceil {q + {i over 2}} rightrceil = q + leftlceil {{i over 2}} rightrceil = q + i cr
& leftlfloor {{{n + 1} over 2}} rightrfloor = leftlfloor {{{2q + 1 + i} over 2}} rightrfloor
= leftlfloor {q + {{1 + i} over 2}} rightrfloor = q + leftlfloor {{{1 + i} over 2}} rightrfloor = q + i cr}
$$
and, in general
$$
leftlceil {{n over m}} rightrceil = leftlfloor {{{n + m - 1} over m}} rightrfloor quad left| matrix{
;n,m in Z hfill cr
;1 le m hfill cr} right.
$$
re. for details to this article.
answered Jan 30 at 12:05
G CabG Cab
20.4k31341
$endgroup$
$begingroup$
How Did you convert the ceiling function to the floor? That is where I stuck actually.
$endgroup$
– Bright Chancellor
Jan 31 at 8:19
1
$begingroup$
@BrightChancellor: put an addendum to answer to this
$endgroup$
– G Cab
Jan 31 at 11:16
$begingroup$
I got it, thanks.
$endgroup$
– Bright Chancellor
Jan 31 at 11:22
add a comment |
$begingroup$
The index of summation is ranging as
$$
a le k le n
$$
Replace it with its even and odd components
$$
k = 2j - iquad left| {,i = 0,1} right.
$$
Then for the even component it shall be
$$
eqalign{
& i = 0quad Rightarrow quad a le 2j le nquad Rightarrow quad leftlceil {{a over 2}} rightrceil le j le leftlfloor {{n over 2}} rightrfloor
quad Rightarrow cr
& Rightarrow quad leftlfloor {{{a + 1} over 2}} rightrfloor le j le leftlfloor {{n over 2}} rightrfloor cr}
$$
and for the odd
$$
eqalign{
& i = 1quad Rightarrow quad a le 2j - 1 le nquad Rightarrow quad leftlceil {{{a + 1} over 2}} rightrceil le j le
leftlfloor {{{n + 1} over 2}} rightrfloor quad Rightarrow cr
& Rightarrow quad leftlfloor {{{a + 2} over 2}} rightrfloor le j le leftlfloor {{{n + 1} over 2}} rightrfloor cr}
$$
Note that for the lower bound we shall employ the ceiling and the floor for the upper. Then we can convert the ceiling to floor.
The sum over $k$ of $f(k)$ can be split into the sum over the even and the odd component
$$
eqalign{
& sumlimits_{a, le ,k, le ;n} {f(k)} = sumlimits_{a, le ,k, le ;n} {left. {f(k),} right|_{,k = 2j} + left. {f(k),} right|_{,k = 2j - 1} } = cr
& = sumlimits_{a, le ,2j, le ;n} {f(2j)} + sumlimits_{a, le ,2j - 1, le ;n} {f(2j - 1)} = cr
& = sumlimits_{leftlfloor {{{a + 1} over 2}} rightrfloor le j le leftlfloor {{n over 2}} rightrfloor } {f(2j)}
+ sumlimits_{leftlfloor {{{a + 2} over 2}} rightrfloor le j le leftlfloor {{{n + 1} over 2}} rightrfloor } {f(2j - 1)} cr}
$$
and same for that of $(-1)^k f(k)$
$$
eqalign{
& sumlimits_{a, le ,k, le ;n} {left( { - 1} right)^{,k} f(k)}
= sumlimits_{leftlfloor {{{a + 1} over 2}} rightrfloor le j le leftlfloor {{n over 2}} rightrfloor } {left( { - 1} right)^{,2j} f(2j)}
+ sumlimits_{leftlfloor {{{a + 2} over 2}} rightrfloor le j le leftlfloor {{{n + 1} over 2}} rightrfloor } {left( { - 1} right)^{,2j - 1} f(2j - 1)} = cr
& = sumlimits_{leftlfloor {{{a + 1} over 2}} rightrfloor le j le leftlfloor {{n over 2}} rightrfloor } {f(2j)}
- sumlimits_{leftlfloor {{{a + 2} over 2}} rightrfloor le j le leftlfloor {{{n + 1} over 2}} rightrfloor } {f(2j - 1)} cr}
$$
Now, just subtract the two above to confirm the thesis.
--- addendum ----
Concerning your request on how to transform ceiling <-> floor, consider that
$$
eqalign{
& n = 2leftlceil {{n over 2}} rightrceil + 2left{ {{n over 2}} right} = 2q + iquad left| matrix{
;n,q in Z hfill cr
,i = 0,1 hfill cr} right. cr
& leftlceil {{n over 2}} rightrceil = leftlceil {{{2q + i} over 2}} rightrceil
= leftlceil {q + {i over 2}} rightrceil = q + leftlceil {{i over 2}} rightrceil = q + i cr
& leftlfloor {{{n + 1} over 2}} rightrfloor = leftlfloor {{{2q + 1 + i} over 2}} rightrfloor
= leftlfloor {q + {{1 + i} over 2}} rightrfloor = q + leftlfloor {{{1 + i} over 2}} rightrfloor = q + i cr}
$$
and, in general
$$
leftlceil {{n over m}} rightrceil = leftlfloor {{{n + m - 1} over m}} rightrfloor quad left| matrix{
;n,m in Z hfill cr
;1 le m hfill cr} right.
$$
re. for details to this article.
answered Jan 30 at 12:05
G CabG Cab
20.4k31341
$endgroup$
The index of summation is ranging as
$$
a le k le n
$$
Replace it with its even and odd components
$$
k = 2j - iquad left| {,i = 0,1} right.
$$
Then for the even component it shall be
$$
eqalign{
& i = 0quad Rightarrow quad a le 2j le nquad Rightarrow quad leftlceil {{a over 2}} rightrceil le j le leftlfloor {{n over 2}} rightrfloor
quad Rightarrow cr
& Rightarrow quad leftlfloor {{{a + 1} over 2}} rightrfloor le j le leftlfloor {{n over 2}} rightrfloor cr}
$$
and for the odd
$$
eqalign{
& i = 1quad Rightarrow quad a le 2j - 1 le nquad Rightarrow quad leftlceil {{{a + 1} over 2}} rightrceil le j le
leftlfloor {{{n + 1} over 2}} rightrfloor quad Rightarrow cr
& Rightarrow quad leftlfloor {{{a + 2} over 2}} rightrfloor le j le leftlfloor {{{n + 1} over 2}} rightrfloor cr}
$$
Note that for the lower bound we shall employ the ceiling and the floor for the upper. Then we can convert the ceiling to floor.
The sum over $k$ of $f(k)$ can be split into the sum over the even and the odd component
$$
eqalign{
& sumlimits_{a, le ,k, le ;n} {f(k)} = sumlimits_{a, le ,k, le ;n} {left. {f(k),} right|_{,k = 2j} + left. {f(k),} right|_{,k = 2j - 1} } = cr
& = sumlimits_{a, le ,2j, le ;n} {f(2j)} + sumlimits_{a, le ,2j - 1, le ;n} {f(2j - 1)} = cr
& = sumlimits_{leftlfloor {{{a + 1} over 2}} rightrfloor le j le leftlfloor {{n over 2}} rightrfloor } {f(2j)}
+ sumlimits_{leftlfloor {{{a + 2} over 2}} rightrfloor le j le leftlfloor {{{n + 1} over 2}} rightrfloor } {f(2j - 1)} cr}
$$
and same for that of $(-1)^k f(k)$
$$
eqalign{
& sumlimits_{a, le ,k, le ;n} {left( { - 1} right)^{,k} f(k)}
= sumlimits_{leftlfloor {{{a + 1} over 2}} rightrfloor le j le leftlfloor {{n over 2}} rightrfloor } {left( { - 1} right)^{,2j} f(2j)}
+ sumlimits_{leftlfloor {{{a + 2} over 2}} rightrfloor le j le leftlfloor {{{n + 1} over 2}} rightrfloor } {left( { - 1} right)^{,2j - 1} f(2j - 1)} = cr
& = sumlimits_{leftlfloor {{{a + 1} over 2}} rightrfloor le j le leftlfloor {{n over 2}} rightrfloor } {f(2j)}
- sumlimits_{leftlfloor {{{a + 2} over 2}} rightrfloor le j le leftlfloor {{{n + 1} over 2}} rightrfloor } {f(2j - 1)} cr}
$$
Now, just subtract the two above to confirm the thesis.
--- addendum ----
Concerning your request on how to transform ceiling <-> floor, consider that
$$
eqalign{
& n = 2leftlceil {{n over 2}} rightrceil + 2left{ {{n over 2}} right} = 2q + iquad left| matrix{
;n,q in Z hfill cr
,i = 0,1 hfill cr} right. cr
& leftlceil {{n over 2}} rightrceil = leftlceil {{{2q + i} over 2}} rightrceil
= leftlceil {q + {i over 2}} rightrceil = q + leftlceil {{i over 2}} rightrceil = q + i cr
& leftlfloor {{{n + 1} over 2}} rightrfloor = leftlfloor {{{2q + 1 + i} over 2}} rightrfloor
= leftlfloor {q + {{1 + i} over 2}} rightrfloor = q + leftlfloor {{{1 + i} over 2}} rightrfloor = q + i cr}
$$
and, in general
$$
leftlceil {{n over m}} rightrceil = leftlfloor {{{n + m - 1} over m}} rightrfloor quad left| matrix{
;n,m in Z hfill cr
;1 le m hfill cr} right.
$$
re. for details to this article.
answered Jan 30 at 12:05
G CabG Cab
20.4k31341
edited Jan 31 at 11:10
answered Jan 30 at 12:05
G CabG Cab
20.4k31341
answered Jan 30 at 12:05
G CabG Cab
20.4k31341
answered Jan 30 at 12:05
G CabG Cab
20.4k31341
20.4k31341
$begingroup$
How Did you convert the ceiling function to the floor? That is where I stuck actually.
$endgroup$
– Bright Chancellor
Jan 31 at 8:19
1
$begingroup$
@BrightChancellor: put an addendum to answer to this
$endgroup$
– G Cab
Jan 31 at 11:16
$begingroup$
I got it, thanks.
$endgroup$
– Bright Chancellor
Jan 31 at 11:22
add a comment |
$begingroup$
How Did you convert the ceiling function to the floor? That is where I stuck actually.
$endgroup$
– Bright Chancellor
Jan 31 at 8:19
1
$begingroup$
@BrightChancellor: put an addendum to answer to this
$endgroup$
– G Cab
Jan 31 at 11:16
$begingroup$
I got it, thanks.
$endgroup$
– Bright Chancellor
Jan 31 at 11:22
$begingroup$
How Did you convert the ceiling function to the floor? That is where I stuck actually.
$endgroup$
– Bright Chancellor
Jan 31 at 8:19
$begingroup$
How Did you convert the ceiling function to the floor? That is where I stuck actually.
$endgroup$
– Bright Chancellor
Jan 31 at 8:19
1
1
$begingroup$
@BrightChancellor: put an addendum to answer to this
$endgroup$
– G Cab
Jan 31 at 11:16
$begingroup$
@BrightChancellor: put an addendum to answer to this
$endgroup$
– G Cab
Jan 31 at 11:16
$begingroup$
I got it, thanks.
$endgroup$
– Bright Chancellor
Jan 31 at 11:22
$begingroup$
I got it, thanks.
$endgroup$
– Bright Chancellor
Jan 31 at 11:22
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%2f3090685%2fa-series-question-whose-indices-involves-greatest-integer-function%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
