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Reducing the white spacing
Reducing the title spacingReducing white space above date in Letter ClassReducing spacing after headingsReducing the spaces before and after smallsetminusNon-uniform white spacing with vspaceReducing chapter and section spacing for scrbookReducing spacing between section headings and text in resumeAutomatically reducing spacing before punctuation in math modereducing the spacing between bib itemsReducing space in the Table of Contents
5
documentclass[11pt, a4paper]report
usepackagebm
usepackageamsfonts, graphicx, verbatim, mathtools,amssymb, amsthm, mathrsfs
usepackagecolor
usepackagearray
usepackagesetspace% if you must (for double spacing thesis)
usepackagefancyhdr
usepackageenumitem
usepackagetikz
usepackageparskip
usepackagelipsum
usepackagefloatrow
begindocument
newcommandiuimkern1mu
beginalign*
setlengthextrarowheight3pt
noindentbegintabular c c c c c
& $0$ & $1$ & $2$ & $3$ & $4$\
cline1-6
$chi_0$ & $1$ & $1$ & $1$ & $1$ & $1$\
$chi_1$ & $1$ & $a$ & $a^2$ & $a^3$ & $a^4$\
$chi_2$ & $1$ & $a^2$ & $a^4$ & $a$ & $a^3$\
$chi_3$ & $1$ & $a^3$ & $a$ & $a^4$ & $a^2$\
$chi_4$ & $1$ & $a^4$ & $a^3$ & $a^2$ & $a$\
endtabular
endalign*
with $a = expfrac2piiu5$ hence $a^5=1$ with $|G|=5$.
Applying the definition of Fourier transform from definition 3.1.2 we have:
doublespacing
$hatf(chi_0)=f(0)+f(1)+f(2)+f(3)+f(4)$\
$hatf(chi_1)=f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4)$\
$hatf(chi_2)=f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4)$\
$hatf(chi_3)=f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4)$\
$hatf(chi_4)=f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)$\
Using definition 3.1.3. we can compute the inverse Fourier transform $f(t)$:
beginalign*
f(0)
&=frac15[ hatf(chi_0)+hatf(chi_1)+hatf(chi_2)+hatf(chi_3)+hatf(chi_4)]\
&beginaligned[t]
=&frac15[f(0)+f(1)+f(2)+f(3)+f(4)]\
+&frac15[f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4)]\
+&frac15[f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4)]\
+&frac15[f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4)]\
+&frac15[f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)]
endaligned
\
&beginaligned[t]
=&f(0)\
+&fracf(1)5[1+a+a^2+a^3+a^4]\
+&fracf(2)5[1+a+a^2+a^3+a^4]\
+&fracf(3)5[1+a+a^2+a^3+a^4]\
+&fracf(4)5[1+a+a^2+a^3+a^4]\
=&f(0)
endaligned
endalign*
Similarly
beginalign*
f(1)
&= frac15[hatf(chi_0)+frac1ahatf(chi_1)+frac1a^2hatf(chi_2)+frac1a^3hatf(chi_3)+frac1a^4hatf(chi_4)]\
&beginaligned[t]
=&f(1)
endaligned
endalign*
beginalign*
f(2)
&= frac15[hatf(chi_0)+a^2hatf(chi_1)+a^4hatf(chi_2)+ahatf(chi_3)+a^3hatf(chi_4)]\
&beginaligned[t]
=&f(2)
endaligned
endalign*
beginalign*
f(3)
&= frac15[hatf(chi_0)+a^3hatf(chi_1)+ahatf(chi_2)+a^4hatf(chi_3)+a^2hatf(chi_4)]\
&beginaligned[t]
=&f(3)
endaligned
endalign*
beginalign*
f(4)
&= frac15[hatf(chi_0)+a^4hatf(chi_1)+a^3hatf(chi_2)+a^2hatf(chi_3)+ahatf(chi_4)]\
&beginaligned[t]
=&f(4)
endaligned
endalign*
enddocument
How can I reduce the spacing in this? Where it says "using definition....." I would like to move this up to $hatf(chi_4)$. Also I want to reduce the spacing of $f(1) = ... = f(1)$ and $f(2) = ... = f(2)$ etc.
Edit: I've attached all 3 pages so you guys get the full picture of what the issue is.
spacing
asked May 1 at 14:25
MathsMaths
47011
add a comment |
5
documentclass[11pt, a4paper]report
usepackagebm
usepackageamsfonts, graphicx, verbatim, mathtools,amssymb, amsthm, mathrsfs
usepackagecolor
usepackagearray
usepackagesetspace% if you must (for double spacing thesis)
usepackagefancyhdr
usepackageenumitem
usepackagetikz
usepackageparskip
usepackagelipsum
usepackagefloatrow
begindocument
newcommandiuimkern1mu
beginalign*
setlengthextrarowheight3pt
noindentbegintabular c c c c c
& $0$ & $1$ & $2$ & $3$ & $4$\
cline1-6
$chi_0$ & $1$ & $1$ & $1$ & $1$ & $1$\
$chi_1$ & $1$ & $a$ & $a^2$ & $a^3$ & $a^4$\
$chi_2$ & $1$ & $a^2$ & $a^4$ & $a$ & $a^3$\
$chi_3$ & $1$ & $a^3$ & $a$ & $a^4$ & $a^2$\
$chi_4$ & $1$ & $a^4$ & $a^3$ & $a^2$ & $a$\
endtabular
endalign*
with $a = expfrac2piiu5$ hence $a^5=1$ with $|G|=5$.
Applying the definition of Fourier transform from definition 3.1.2 we have:
doublespacing
$hatf(chi_0)=f(0)+f(1)+f(2)+f(3)+f(4)$\
$hatf(chi_1)=f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4)$\
$hatf(chi_2)=f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4)$\
$hatf(chi_3)=f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4)$\
$hatf(chi_4)=f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)$\
Using definition 3.1.3. we can compute the inverse Fourier transform $f(t)$:
beginalign*
f(0)
&=frac15[ hatf(chi_0)+hatf(chi_1)+hatf(chi_2)+hatf(chi_3)+hatf(chi_4)]\
&beginaligned[t]
=&frac15[f(0)+f(1)+f(2)+f(3)+f(4)]\
+&frac15[f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4)]\
+&frac15[f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4)]\
+&frac15[f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4)]\
+&frac15[f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)]
endaligned
\
&beginaligned[t]
=&f(0)\
+&fracf(1)5[1+a+a^2+a^3+a^4]\
+&fracf(2)5[1+a+a^2+a^3+a^4]\
+&fracf(3)5[1+a+a^2+a^3+a^4]\
+&fracf(4)5[1+a+a^2+a^3+a^4]\
=&f(0)
endaligned
endalign*
Similarly
beginalign*
f(1)
&= frac15[hatf(chi_0)+frac1ahatf(chi_1)+frac1a^2hatf(chi_2)+frac1a^3hatf(chi_3)+frac1a^4hatf(chi_4)]\
&beginaligned[t]
=&f(1)
endaligned
endalign*
beginalign*
f(2)
&= frac15[hatf(chi_0)+a^2hatf(chi_1)+a^4hatf(chi_2)+ahatf(chi_3)+a^3hatf(chi_4)]\
&beginaligned[t]
=&f(2)
endaligned
endalign*
beginalign*
f(3)
&= frac15[hatf(chi_0)+a^3hatf(chi_1)+ahatf(chi_2)+a^4hatf(chi_3)+a^2hatf(chi_4)]\
&beginaligned[t]
=&f(3)
endaligned
endalign*
beginalign*
f(4)
&= frac15[hatf(chi_0)+a^4hatf(chi_1)+a^3hatf(chi_2)+a^2hatf(chi_3)+ahatf(chi_4)]\
&beginaligned[t]
=&f(4)
endaligned
endalign*
enddocument
How can I reduce the spacing in this? Where it says "using definition....." I would like to move this up to $hatf(chi_4)$. Also I want to reduce the spacing of $f(1) = ... = f(1)$ and $f(2) = ... = f(2)$ etc.
Edit: I've attached all 3 pages so you guys get the full picture of what the issue is.
spacing
asked May 1 at 14:25
MathsMaths
47011
Unrelated to the issue, but switching from tabular to array you can remove all the repeated $ signs.
– leandriis
May 1 at 16:37
add a comment |
5
5
5
1
documentclass[11pt, a4paper]report
usepackagebm
usepackageamsfonts, graphicx, verbatim, mathtools,amssymb, amsthm, mathrsfs
usepackagecolor
usepackagearray
usepackagesetspace% if you must (for double spacing thesis)
usepackagefancyhdr
usepackageenumitem
usepackagetikz
usepackageparskip
usepackagelipsum
usepackagefloatrow
begindocument
newcommandiuimkern1mu
beginalign*
setlengthextrarowheight3pt
noindentbegintabular c c c c c
& $0$ & $1$ & $2$ & $3$ & $4$\
cline1-6
$chi_0$ & $1$ & $1$ & $1$ & $1$ & $1$\
$chi_1$ & $1$ & $a$ & $a^2$ & $a^3$ & $a^4$\
$chi_2$ & $1$ & $a^2$ & $a^4$ & $a$ & $a^3$\
$chi_3$ & $1$ & $a^3$ & $a$ & $a^4$ & $a^2$\
$chi_4$ & $1$ & $a^4$ & $a^3$ & $a^2$ & $a$\
endtabular
endalign*
with $a = expfrac2piiu5$ hence $a^5=1$ with $|G|=5$.
Applying the definition of Fourier transform from definition 3.1.2 we have:
doublespacing
$hatf(chi_0)=f(0)+f(1)+f(2)+f(3)+f(4)$\
$hatf(chi_1)=f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4)$\
$hatf(chi_2)=f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4)$\
$hatf(chi_3)=f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4)$\
$hatf(chi_4)=f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)$\
Using definition 3.1.3. we can compute the inverse Fourier transform $f(t)$:
beginalign*
f(0)
&=frac15[ hatf(chi_0)+hatf(chi_1)+hatf(chi_2)+hatf(chi_3)+hatf(chi_4)]\
&beginaligned[t]
=&frac15[f(0)+f(1)+f(2)+f(3)+f(4)]\
+&frac15[f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4)]\
+&frac15[f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4)]\
+&frac15[f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4)]\
+&frac15[f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)]
endaligned
\
&beginaligned[t]
=&f(0)\
+&fracf(1)5[1+a+a^2+a^3+a^4]\
+&fracf(2)5[1+a+a^2+a^3+a^4]\
+&fracf(3)5[1+a+a^2+a^3+a^4]\
+&fracf(4)5[1+a+a^2+a^3+a^4]\
=&f(0)
endaligned
endalign*
Similarly
beginalign*
f(1)
&= frac15[hatf(chi_0)+frac1ahatf(chi_1)+frac1a^2hatf(chi_2)+frac1a^3hatf(chi_3)+frac1a^4hatf(chi_4)]\
&beginaligned[t]
=&f(1)
endaligned
endalign*
beginalign*
f(2)
&= frac15[hatf(chi_0)+a^2hatf(chi_1)+a^4hatf(chi_2)+ahatf(chi_3)+a^3hatf(chi_4)]\
&beginaligned[t]
=&f(2)
endaligned
endalign*
beginalign*
f(3)
&= frac15[hatf(chi_0)+a^3hatf(chi_1)+ahatf(chi_2)+a^4hatf(chi_3)+a^2hatf(chi_4)]\
&beginaligned[t]
=&f(3)
endaligned
endalign*
beginalign*
f(4)
&= frac15[hatf(chi_0)+a^4hatf(chi_1)+a^3hatf(chi_2)+a^2hatf(chi_3)+ahatf(chi_4)]\
&beginaligned[t]
=&f(4)
endaligned
endalign*
enddocument
How can I reduce the spacing in this? Where it says "using definition....." I would like to move this up to $hatf(chi_4)$. Also I want to reduce the spacing of $f(1) = ... = f(1)$ and $f(2) = ... = f(2)$ etc.
Edit: I've attached all 3 pages so you guys get the full picture of what the issue is.
spacing
asked May 1 at 14:25
MathsMaths
47011
documentclass[11pt, a4paper]report
usepackagebm
usepackageamsfonts, graphicx, verbatim, mathtools,amssymb, amsthm, mathrsfs
usepackagecolor
usepackagearray
usepackagesetspace% if you must (for double spacing thesis)
usepackagefancyhdr
usepackageenumitem
usepackagetikz
usepackageparskip
usepackagelipsum
usepackagefloatrow
begindocument
newcommandiuimkern1mu
beginalign*
setlengthextrarowheight3pt
noindentbegintabular c c c c c
& $0$ & $1$ & $2$ & $3$ & $4$\
cline1-6
$chi_0$ & $1$ & $1$ & $1$ & $1$ & $1$\
$chi_1$ & $1$ & $a$ & $a^2$ & $a^3$ & $a^4$\
$chi_2$ & $1$ & $a^2$ & $a^4$ & $a$ & $a^3$\
$chi_3$ & $1$ & $a^3$ & $a$ & $a^4$ & $a^2$\
$chi_4$ & $1$ & $a^4$ & $a^3$ & $a^2$ & $a$\
endtabular
endalign*
with $a = expfrac2piiu5$ hence $a^5=1$ with $|G|=5$.
Applying the definition of Fourier transform from definition 3.1.2 we have:
doublespacing
$hatf(chi_0)=f(0)+f(1)+f(2)+f(3)+f(4)$\
$hatf(chi_1)=f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4)$\
$hatf(chi_2)=f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4)$\
$hatf(chi_3)=f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4)$\
$hatf(chi_4)=f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)$\
Using definition 3.1.3. we can compute the inverse Fourier transform $f(t)$:
beginalign*
f(0)
&=frac15[ hatf(chi_0)+hatf(chi_1)+hatf(chi_2)+hatf(chi_3)+hatf(chi_4)]\
&beginaligned[t]
=&frac15[f(0)+f(1)+f(2)+f(3)+f(4)]\
+&frac15[f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4)]\
+&frac15[f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4)]\
+&frac15[f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4)]\
+&frac15[f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)]
endaligned
\
&beginaligned[t]
=&f(0)\
+&fracf(1)5[1+a+a^2+a^3+a^4]\
+&fracf(2)5[1+a+a^2+a^3+a^4]\
+&fracf(3)5[1+a+a^2+a^3+a^4]\
+&fracf(4)5[1+a+a^2+a^3+a^4]\
=&f(0)
endaligned
endalign*
Similarly
beginalign*
f(1)
&= frac15[hatf(chi_0)+frac1ahatf(chi_1)+frac1a^2hatf(chi_2)+frac1a^3hatf(chi_3)+frac1a^4hatf(chi_4)]\
&beginaligned[t]
=&f(1)
endaligned
endalign*
beginalign*
f(2)
&= frac15[hatf(chi_0)+a^2hatf(chi_1)+a^4hatf(chi_2)+ahatf(chi_3)+a^3hatf(chi_4)]\
&beginaligned[t]
=&f(2)
endaligned
endalign*
beginalign*
f(3)
&= frac15[hatf(chi_0)+a^3hatf(chi_1)+ahatf(chi_2)+a^4hatf(chi_3)+a^2hatf(chi_4)]\
&beginaligned[t]
=&f(3)
endaligned
endalign*
beginalign*
f(4)
&= frac15[hatf(chi_0)+a^4hatf(chi_1)+a^3hatf(chi_2)+a^2hatf(chi_3)+ahatf(chi_4)]\
&beginaligned[t]
=&f(4)
endaligned
endalign*
enddocument
How can I reduce the spacing in this? Where it says "using definition....." I would like to move this up to $hatf(chi_4)$. Also I want to reduce the spacing of $f(1) = ... = f(1)$ and $f(2) = ... = f(2)$ etc.
Edit: I've attached all 3 pages so you guys get the full picture of what the issue is.
spacing
spacing
asked May 1 at 14:25
MathsMaths
47011
asked May 1 at 14:25
MathsMaths
47011
edited May 1 at 14:53
Maths
asked May 1 at 14:25
MathsMaths
47011
asked May 1 at 14:25
MathsMaths
47011
asked May 1 at 14:25
MathsMaths
47011
47011
Unrelated to the issue, but switching from tabular to array you can remove all the repeated $ signs.
– leandriis
May 1 at 16:37
add a comment |
Unrelated to the issue, but switching from tabular to array you can remove all the repeated $ signs.
– leandriis
May 1 at 16:37
Unrelated to the issue, but switching from tabular to array you can remove all the repeated $ signs.
– leandriis
May 1 at 16:37
Unrelated to the issue, but switching from tabular to array you can remove all the repeated $ signs.
– leandriis
May 1 at 16:37
add a comment |
2 Answers
2
active
oldest
votes
4
With this simpler code, it can all fit on a single page. I loaded nccmath for its medium-sized fractions, which look better for coefficients, in my opinion:
documentclass[11pt, a4paper]report
usepackage[utf8]inputenc
usepackage[T1]fontenc
usepackagebm
usepackagenccmath
usepackageamsfonts, graphicx, verbatim, mathtools,amssymb, amsthm, mathrsfs
usepackagecolor
usepackagearray
usepackagesetspace% if you must (for double spacing thesis)
usepackagefancyhdr
usepackageenumitem
usepackagetikz
usepackageparskip
usepackagelipsum
usepackagefloatrow
begindocument
newcommandiuimkern1mu
[
setlengthextrarowheight3pt
beginarray c c c c c
& 0 & 1 & 2 & 3 & 4\
cline1-6
chi_0 & 1 & 1 & 1 & 1 & 1\
chi_1 & 1 & a & a^2 & a^3 & a^4\
chi_2 & 1 & a^2 & a^4 & a & a^3\
chi_3 & 1 & a^3 & a & a^4 & a^2\
chi_4 & 1 & a^4 & a^3 & a^2 & a\
endarray
]
with $a = expbiglfrac2pi iu5bigr$, hence $a^5=1$ with $|G|=5$.
Applying the definition of Fourier transform from definition 3.1.2 we have:
beginfleqn
beginalign*
hatf(chi_0) & =f(0)+f(1)+f(2)+f(3)+f(4) \
hatf(chi_1) & =f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4) \
hatf(chi_2) & =f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4) \
hatf(chi_3) & =f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4) \
hatf(chi_4) & =f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)
endalign*
endfleqn
Using definition 3.1.3. we can compute the inverse Fourier transform $f(t)$:
allowdisplaybreaks
beginalign*
f(0)
&=mfrac15bigl[ hatf(chi_0)+hatf(chi_1)+hatf(chi_2)+hatf(chi_3)+hatf(chi_4)bigr]\
& = beginaligned[t]
&mfrac15bigl[f(0)+f(1)+f(2)+f(3)+f(4)]\
& + mfrac15bigl[f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4)bigr]\
& + mfrac15bigl[f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4)bigr]\
& + mfrac15bigl[f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4)bigr]\
& + mfrac15bigl[f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)bigr]
endaligned\
& =f(0) beginaligned[t]
& + mfracf(1)5[1+a+a^2+a^3+a^4]\
& + mfracf(2)5[1+a+a^2+a^3+a^4]\
& + mfracf(3)5[1+a+a^2+a^3+a^4]\
& + mfracf(4)5[1+a+a^2+a^3+a^4]
endaligned\
& = f(0)
shortintertextSimilarly:
f(1)
&= mfrac15Bigl[hatf(chi_0)+mfrac1ahatf(chi_1)+mfrac1a^2hatf(chi_2)+mfrac1a^3hatf(chi_3)+mfrac1a^4hatf(chi_4)Bigr]\
& = f(1) \[1.5ex]
f(2)
&= mfrac15bigl[hatf(chi_0)+a^2hatf(chi_1)+a^4hatf(chi_2)+ahatf(chi_3)+a^3hatf(chi_4)bigr] \
& = f(2) \[1.5ex]
f(3)
&= mfrac15bigl[hatf(chi_0)+a^3hatf(chi_1)+ahatf(chi_2)+a^4hatf(chi_3)+a^2hatf(chi_4)bigr] \
& = f(3) \[1.5ex]
f(4)
&= mfrac15bigl[hatf(chi_0)+a^4hatf(chi_1)+a^3hatf(chi_2)+a^2hatf(chi_3)+ahatf(chi_4)bigr] \
& = f(4)
endalign*
enddocument
answered May 1 at 16:54
BernardBernard
178k779211
this is nice however you missed out the remaining section for $f(0)$ aha!
– Maths
2 days ago
Oh! yes. I'll fix it in a moment
– Bernard
2 days ago
I had to slightly modify the code to make it fit on a single page (replacedintertextwithshortintertext, and loadingnccmathbeforemathtoolsto make it work).
– Bernard
2 days ago
there's still some code of f(0) missing aha. its where I group f(1) ... f(4) as f(1)[1+a+...+a^4] etc. Also, I don't mind if it runs over two pages. I don't want the text to be squashed, all I wanted is to make use of the empty white space :)
– Maths
2 days ago
Refer to my code in the question, you'll see the part you missed :)
– Maths
2 days ago
|
show 5 more comments
4
You should avoid \ on the last line of alignments. Perhaps the following is closer to what you want:
documentclass[11pt, a4paper]report
usepackageamsmath,array
begindocument
newcommandiuimkern1mu
beginequation*
setlengthextrarowheight3pt
begintabular c c c c c
& $0$ & $1$ & $2$ & $3$ & $4$\
cline1-6
$chi_0$ & $1$ & $1$ & $1$ & $1$ & $1$\
$chi_1$ & $1$ & $a$ & $a^2$ & $a^3$ & $a^4$\
$chi_2$ & $1$ & $a^2$ & $a^4$ & $a$ & $a^3$\
$chi_3$ & $1$ & $a^3$ & $a$ & $a^4$ & $a^2$\
$chi_4$ & $1$ & $a^4$ & $a^3$ & $a^2$ & $a$\
endtabular
endequation*
with $a = expfrac2piiu5$ hence $a^5=1$ with $|G|=5$.
Applying the definition of Fourier transform from Definition~3.1.2 we
have:
beginalign*
hatf(chi_0) &=f(0)+f(1)+f(2)+f(3)+f(4),\
hatf(chi_1) &=f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4),\
hatf(chi_2) &=f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4),\
hatf(chi_3) &=f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4),\
hatf(chi_4) &=f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4).
endalign*
Using Definition~3.1.3 we can compute the inverse Fourier transform
$f(t)$:
beginalign*
f(0)
&=frac15[ hatf(chi_0) + hatf(chi_1) + hatf(chi_2) +
hatf(chi_3) + hatf(chi_4)]\
&=frac15[f(0)+f(1)+f(2)+f(3)+f(4)]\
&qquad + frac15[f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4)]\
&qquad + frac15[f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4)]\
&qquad + frac15[f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4)]\
&qquad + frac15[f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)]
\
&= f(0)\
&qquad + fracf(1)5[1+a+a^2+a^3+a^4]\
&qquad +fracf(2)5[1+a+a^2+a^3+a^4]\
&qquad +fracf(3)5[1+a+a^2+a^3+a^4]\
&qquad +fracf(4)5[1+a+a^2+a^3+a^4]\
&=f(0).
endalign*
Similarly
beginalign*
f(1)
&= frac15Bigl[hatf(chi_0) + frac1ahatf(chi_1) +
frac1a^2hatf(chi_2) + frac1a^3hatf(chi_3) +
frac1a^4hatf(chi_4)Bigr]\
&=f(1),\
f(2)
&= frac15[hatf(chi_0) + a^2hatf(chi_1) +
a^4hatf(chi_2) + ahatf(chi_3) + a^3hatf(chi_4)]\
&=f(2), \
f(3)
&= frac15[hatf(chi_0) + a^3hatf(chi_1) +
ahatf(chi_2) + a^4hatf(chi_3) + a^2hatf(chi_4)]\
&=f(3),\
f(4)
&= frac15[hatf(chi_0) + a^4hatf(chi_1) +
a^3hatf(chi_2) + a^2hatf(chi_3) + ahatf(chi_4)]\
& =f(4).
endalign*
enddocument
answered May 1 at 15:09
Andrew SwannAndrew Swann
78.8k9138336
why did you push f(0) outwards? it wasn't necessary. but thanks for your solution
– Maths
May 1 at 15:42
1
The+'s should not be under the=, but to the right of it as they belong to that side of the equation. Whether you want to indent byqquadas I did, or the smallerquadis a matter of taste.
– Andrew Swann
May 1 at 18:28
add a comment |
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4
With this simpler code, it can all fit on a single page. I loaded nccmath for its medium-sized fractions, which look better for coefficients, in my opinion:
documentclass[11pt, a4paper]report
usepackage[utf8]inputenc
usepackage[T1]fontenc
usepackagebm
usepackagenccmath
usepackageamsfonts, graphicx, verbatim, mathtools,amssymb, amsthm, mathrsfs
usepackagecolor
usepackagearray
usepackagesetspace% if you must (for double spacing thesis)
usepackagefancyhdr
usepackageenumitem
usepackagetikz
usepackageparskip
usepackagelipsum
usepackagefloatrow
begindocument
newcommandiuimkern1mu
[
setlengthextrarowheight3pt
beginarray c c c c c
& 0 & 1 & 2 & 3 & 4\
cline1-6
chi_0 & 1 & 1 & 1 & 1 & 1\
chi_1 & 1 & a & a^2 & a^3 & a^4\
chi_2 & 1 & a^2 & a^4 & a & a^3\
chi_3 & 1 & a^3 & a & a^4 & a^2\
chi_4 & 1 & a^4 & a^3 & a^2 & a\
endarray
]
with $a = expbiglfrac2pi iu5bigr$, hence $a^5=1$ with $|G|=5$.
Applying the definition of Fourier transform from definition 3.1.2 we have:
beginfleqn
beginalign*
hatf(chi_0) & =f(0)+f(1)+f(2)+f(3)+f(4) \
hatf(chi_1) & =f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4) \
hatf(chi_2) & =f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4) \
hatf(chi_3) & =f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4) \
hatf(chi_4) & =f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)
endalign*
endfleqn
Using definition 3.1.3. we can compute the inverse Fourier transform $f(t)$:
allowdisplaybreaks
beginalign*
f(0)
&=mfrac15bigl[ hatf(chi_0)+hatf(chi_1)+hatf(chi_2)+hatf(chi_3)+hatf(chi_4)bigr]\
& = beginaligned[t]
&mfrac15bigl[f(0)+f(1)+f(2)+f(3)+f(4)]\
& + mfrac15bigl[f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4)bigr]\
& + mfrac15bigl[f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4)bigr]\
& + mfrac15bigl[f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4)bigr]\
& + mfrac15bigl[f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)bigr]
endaligned\
& =f(0) beginaligned[t]
& + mfracf(1)5[1+a+a^2+a^3+a^4]\
& + mfracf(2)5[1+a+a^2+a^3+a^4]\
& + mfracf(3)5[1+a+a^2+a^3+a^4]\
& + mfracf(4)5[1+a+a^2+a^3+a^4]
endaligned\
& = f(0)
shortintertextSimilarly:
f(1)
&= mfrac15Bigl[hatf(chi_0)+mfrac1ahatf(chi_1)+mfrac1a^2hatf(chi_2)+mfrac1a^3hatf(chi_3)+mfrac1a^4hatf(chi_4)Bigr]\
& = f(1) \[1.5ex]
f(2)
&= mfrac15bigl[hatf(chi_0)+a^2hatf(chi_1)+a^4hatf(chi_2)+ahatf(chi_3)+a^3hatf(chi_4)bigr] \
& = f(2) \[1.5ex]
f(3)
&= mfrac15bigl[hatf(chi_0)+a^3hatf(chi_1)+ahatf(chi_2)+a^4hatf(chi_3)+a^2hatf(chi_4)bigr] \
& = f(3) \[1.5ex]
f(4)
&= mfrac15bigl[hatf(chi_0)+a^4hatf(chi_1)+a^3hatf(chi_2)+a^2hatf(chi_3)+ahatf(chi_4)bigr] \
& = f(4)
endalign*
enddocument
answered May 1 at 16:54
BernardBernard
178k779211
this is nice however you missed out the remaining section for $f(0)$ aha!
– Maths
2 days ago
Oh! yes. I'll fix it in a moment
– Bernard
2 days ago
I had to slightly modify the code to make it fit on a single page (replacedintertextwithshortintertext, and loadingnccmathbeforemathtoolsto make it work).
– Bernard
2 days ago
there's still some code of f(0) missing aha. its where I group f(1) ... f(4) as f(1)[1+a+...+a^4] etc. Also, I don't mind if it runs over two pages. I don't want the text to be squashed, all I wanted is to make use of the empty white space :)
– Maths
2 days ago
Refer to my code in the question, you'll see the part you missed :)
– Maths
2 days ago
|
show 5 more comments
4
With this simpler code, it can all fit on a single page. I loaded nccmath for its medium-sized fractions, which look better for coefficients, in my opinion:
documentclass[11pt, a4paper]report
usepackage[utf8]inputenc
usepackage[T1]fontenc
usepackagebm
usepackagenccmath
usepackageamsfonts, graphicx, verbatim, mathtools,amssymb, amsthm, mathrsfs
usepackagecolor
usepackagearray
usepackagesetspace% if you must (for double spacing thesis)
usepackagefancyhdr
usepackageenumitem
usepackagetikz
usepackageparskip
usepackagelipsum
usepackagefloatrow
begindocument
newcommandiuimkern1mu
[
setlengthextrarowheight3pt
beginarray c c c c c
& 0 & 1 & 2 & 3 & 4\
cline1-6
chi_0 & 1 & 1 & 1 & 1 & 1\
chi_1 & 1 & a & a^2 & a^3 & a^4\
chi_2 & 1 & a^2 & a^4 & a & a^3\
chi_3 & 1 & a^3 & a & a^4 & a^2\
chi_4 & 1 & a^4 & a^3 & a^2 & a\
endarray
]
with $a = expbiglfrac2pi iu5bigr$, hence $a^5=1$ with $|G|=5$.
Applying the definition of Fourier transform from definition 3.1.2 we have:
beginfleqn
beginalign*
hatf(chi_0) & =f(0)+f(1)+f(2)+f(3)+f(4) \
hatf(chi_1) & =f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4) \
hatf(chi_2) & =f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4) \
hatf(chi_3) & =f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4) \
hatf(chi_4) & =f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)
endalign*
endfleqn
Using definition 3.1.3. we can compute the inverse Fourier transform $f(t)$:
allowdisplaybreaks
beginalign*
f(0)
&=mfrac15bigl[ hatf(chi_0)+hatf(chi_1)+hatf(chi_2)+hatf(chi_3)+hatf(chi_4)bigr]\
& = beginaligned[t]
&mfrac15bigl[f(0)+f(1)+f(2)+f(3)+f(4)]\
& + mfrac15bigl[f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4)bigr]\
& + mfrac15bigl[f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4)bigr]\
& + mfrac15bigl[f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4)bigr]\
& + mfrac15bigl[f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)bigr]
endaligned\
& =f(0) beginaligned[t]
& + mfracf(1)5[1+a+a^2+a^3+a^4]\
& + mfracf(2)5[1+a+a^2+a^3+a^4]\
& + mfracf(3)5[1+a+a^2+a^3+a^4]\
& + mfracf(4)5[1+a+a^2+a^3+a^4]
endaligned\
& = f(0)
shortintertextSimilarly:
f(1)
&= mfrac15Bigl[hatf(chi_0)+mfrac1ahatf(chi_1)+mfrac1a^2hatf(chi_2)+mfrac1a^3hatf(chi_3)+mfrac1a^4hatf(chi_4)Bigr]\
& = f(1) \[1.5ex]
f(2)
&= mfrac15bigl[hatf(chi_0)+a^2hatf(chi_1)+a^4hatf(chi_2)+ahatf(chi_3)+a^3hatf(chi_4)bigr] \
& = f(2) \[1.5ex]
f(3)
&= mfrac15bigl[hatf(chi_0)+a^3hatf(chi_1)+ahatf(chi_2)+a^4hatf(chi_3)+a^2hatf(chi_4)bigr] \
& = f(3) \[1.5ex]
f(4)
&= mfrac15bigl[hatf(chi_0)+a^4hatf(chi_1)+a^3hatf(chi_2)+a^2hatf(chi_3)+ahatf(chi_4)bigr] \
& = f(4)
endalign*
enddocument
answered May 1 at 16:54
BernardBernard
178k779211
this is nice however you missed out the remaining section for $f(0)$ aha!
– Maths
2 days ago
Oh! yes. I'll fix it in a moment
– Bernard
2 days ago
I had to slightly modify the code to make it fit on a single page (replacedintertextwithshortintertext, and loadingnccmathbeforemathtoolsto make it work).
– Bernard
2 days ago
there's still some code of f(0) missing aha. its where I group f(1) ... f(4) as f(1)[1+a+...+a^4] etc. Also, I don't mind if it runs over two pages. I don't want the text to be squashed, all I wanted is to make use of the empty white space :)
– Maths
2 days ago
Refer to my code in the question, you'll see the part you missed :)
– Maths
2 days ago
|
show 5 more comments
4
4
4
With this simpler code, it can all fit on a single page. I loaded nccmath for its medium-sized fractions, which look better for coefficients, in my opinion:
documentclass[11pt, a4paper]report
usepackage[utf8]inputenc
usepackage[T1]fontenc
usepackagebm
usepackagenccmath
usepackageamsfonts, graphicx, verbatim, mathtools,amssymb, amsthm, mathrsfs
usepackagecolor
usepackagearray
usepackagesetspace% if you must (for double spacing thesis)
usepackagefancyhdr
usepackageenumitem
usepackagetikz
usepackageparskip
usepackagelipsum
usepackagefloatrow
begindocument
newcommandiuimkern1mu
[
setlengthextrarowheight3pt
beginarray c c c c c
& 0 & 1 & 2 & 3 & 4\
cline1-6
chi_0 & 1 & 1 & 1 & 1 & 1\
chi_1 & 1 & a & a^2 & a^3 & a^4\
chi_2 & 1 & a^2 & a^4 & a & a^3\
chi_3 & 1 & a^3 & a & a^4 & a^2\
chi_4 & 1 & a^4 & a^3 & a^2 & a\
endarray
]
with $a = expbiglfrac2pi iu5bigr$, hence $a^5=1$ with $|G|=5$.
Applying the definition of Fourier transform from definition 3.1.2 we have:
beginfleqn
beginalign*
hatf(chi_0) & =f(0)+f(1)+f(2)+f(3)+f(4) \
hatf(chi_1) & =f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4) \
hatf(chi_2) & =f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4) \
hatf(chi_3) & =f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4) \
hatf(chi_4) & =f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)
endalign*
endfleqn
Using definition 3.1.3. we can compute the inverse Fourier transform $f(t)$:
allowdisplaybreaks
beginalign*
f(0)
&=mfrac15bigl[ hatf(chi_0)+hatf(chi_1)+hatf(chi_2)+hatf(chi_3)+hatf(chi_4)bigr]\
& = beginaligned[t]
&mfrac15bigl[f(0)+f(1)+f(2)+f(3)+f(4)]\
& + mfrac15bigl[f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4)bigr]\
& + mfrac15bigl[f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4)bigr]\
& + mfrac15bigl[f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4)bigr]\
& + mfrac15bigl[f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)bigr]
endaligned\
& =f(0) beginaligned[t]
& + mfracf(1)5[1+a+a^2+a^3+a^4]\
& + mfracf(2)5[1+a+a^2+a^3+a^4]\
& + mfracf(3)5[1+a+a^2+a^3+a^4]\
& + mfracf(4)5[1+a+a^2+a^3+a^4]
endaligned\
& = f(0)
shortintertextSimilarly:
f(1)
&= mfrac15Bigl[hatf(chi_0)+mfrac1ahatf(chi_1)+mfrac1a^2hatf(chi_2)+mfrac1a^3hatf(chi_3)+mfrac1a^4hatf(chi_4)Bigr]\
& = f(1) \[1.5ex]
f(2)
&= mfrac15bigl[hatf(chi_0)+a^2hatf(chi_1)+a^4hatf(chi_2)+ahatf(chi_3)+a^3hatf(chi_4)bigr] \
& = f(2) \[1.5ex]
f(3)
&= mfrac15bigl[hatf(chi_0)+a^3hatf(chi_1)+ahatf(chi_2)+a^4hatf(chi_3)+a^2hatf(chi_4)bigr] \
& = f(3) \[1.5ex]
f(4)
&= mfrac15bigl[hatf(chi_0)+a^4hatf(chi_1)+a^3hatf(chi_2)+a^2hatf(chi_3)+ahatf(chi_4)bigr] \
& = f(4)
endalign*
enddocument
answered May 1 at 16:54
BernardBernard
178k779211
With this simpler code, it can all fit on a single page. I loaded nccmath for its medium-sized fractions, which look better for coefficients, in my opinion:
documentclass[11pt, a4paper]report
usepackage[utf8]inputenc
usepackage[T1]fontenc
usepackagebm
usepackagenccmath
usepackageamsfonts, graphicx, verbatim, mathtools,amssymb, amsthm, mathrsfs
usepackagecolor
usepackagearray
usepackagesetspace% if you must (for double spacing thesis)
usepackagefancyhdr
usepackageenumitem
usepackagetikz
usepackageparskip
usepackagelipsum
usepackagefloatrow
begindocument
newcommandiuimkern1mu
[
setlengthextrarowheight3pt
beginarray c c c c c
& 0 & 1 & 2 & 3 & 4\
cline1-6
chi_0 & 1 & 1 & 1 & 1 & 1\
chi_1 & 1 & a & a^2 & a^3 & a^4\
chi_2 & 1 & a^2 & a^4 & a & a^3\
chi_3 & 1 & a^3 & a & a^4 & a^2\
chi_4 & 1 & a^4 & a^3 & a^2 & a\
endarray
]
with $a = expbiglfrac2pi iu5bigr$, hence $a^5=1$ with $|G|=5$.
Applying the definition of Fourier transform from definition 3.1.2 we have:
beginfleqn
beginalign*
hatf(chi_0) & =f(0)+f(1)+f(2)+f(3)+f(4) \
hatf(chi_1) & =f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4) \
hatf(chi_2) & =f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4) \
hatf(chi_3) & =f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4) \
hatf(chi_4) & =f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)
endalign*
endfleqn
Using definition 3.1.3. we can compute the inverse Fourier transform $f(t)$:
allowdisplaybreaks
beginalign*
f(0)
&=mfrac15bigl[ hatf(chi_0)+hatf(chi_1)+hatf(chi_2)+hatf(chi_3)+hatf(chi_4)bigr]\
& = beginaligned[t]
&mfrac15bigl[f(0)+f(1)+f(2)+f(3)+f(4)]\
& + mfrac15bigl[f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4)bigr]\
& + mfrac15bigl[f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4)bigr]\
& + mfrac15bigl[f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4)bigr]\
& + mfrac15bigl[f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)bigr]
endaligned\
& =f(0) beginaligned[t]
& + mfracf(1)5[1+a+a^2+a^3+a^4]\
& + mfracf(2)5[1+a+a^2+a^3+a^4]\
& + mfracf(3)5[1+a+a^2+a^3+a^4]\
& + mfracf(4)5[1+a+a^2+a^3+a^4]
endaligned\
& = f(0)
shortintertextSimilarly:
f(1)
&= mfrac15Bigl[hatf(chi_0)+mfrac1ahatf(chi_1)+mfrac1a^2hatf(chi_2)+mfrac1a^3hatf(chi_3)+mfrac1a^4hatf(chi_4)Bigr]\
& = f(1) \[1.5ex]
f(2)
&= mfrac15bigl[hatf(chi_0)+a^2hatf(chi_1)+a^4hatf(chi_2)+ahatf(chi_3)+a^3hatf(chi_4)bigr] \
& = f(2) \[1.5ex]
f(3)
&= mfrac15bigl[hatf(chi_0)+a^3hatf(chi_1)+ahatf(chi_2)+a^4hatf(chi_3)+a^2hatf(chi_4)bigr] \
& = f(3) \[1.5ex]
f(4)
&= mfrac15bigl[hatf(chi_0)+a^4hatf(chi_1)+a^3hatf(chi_2)+a^2hatf(chi_3)+ahatf(chi_4)bigr] \
& = f(4)
endalign*
enddocument
answered May 1 at 16:54
BernardBernard
178k779211
edited 2 days ago
answered May 1 at 16:54
BernardBernard
178k779211
answered May 1 at 16:54
BernardBernard
178k779211
answered May 1 at 16:54
BernardBernard
178k779211
178k779211
this is nice however you missed out the remaining section for $f(0)$ aha!
– Maths
2 days ago
Oh! yes. I'll fix it in a moment
– Bernard
2 days ago
I had to slightly modify the code to make it fit on a single page (replacedintertextwithshortintertext, and loadingnccmathbeforemathtoolsto make it work).
– Bernard
2 days ago
there's still some code of f(0) missing aha. its where I group f(1) ... f(4) as f(1)[1+a+...+a^4] etc. Also, I don't mind if it runs over two pages. I don't want the text to be squashed, all I wanted is to make use of the empty white space :)
– Maths
2 days ago
Refer to my code in the question, you'll see the part you missed :)
– Maths
2 days ago
|
show 5 more comments
this is nice however you missed out the remaining section for $f(0)$ aha!
– Maths
2 days ago
Oh! yes. I'll fix it in a moment
– Bernard
2 days ago
I had to slightly modify the code to make it fit on a single page (replacedintertextwithshortintertext, and loadingnccmathbeforemathtoolsto make it work).
– Bernard
2 days ago
there's still some code of f(0) missing aha. its where I group f(1) ... f(4) as f(1)[1+a+...+a^4] etc. Also, I don't mind if it runs over two pages. I don't want the text to be squashed, all I wanted is to make use of the empty white space :)
– Maths
2 days ago
Refer to my code in the question, you'll see the part you missed :)
– Maths
2 days ago
this is nice however you missed out the remaining section for $f(0)$ aha!
– Maths
2 days ago
this is nice however you missed out the remaining section for $f(0)$ aha!
– Maths
2 days ago
Oh! yes. I'll fix it in a moment
– Bernard
2 days ago
Oh! yes. I'll fix it in a moment
– Bernard
2 days ago
I had to slightly modify the code to make it fit on a single page (replaced
intertext with shortintertext, and loading nccmath before mathtools to make it work).– Bernard
2 days ago
I had to slightly modify the code to make it fit on a single page (replaced
intertext with shortintertext, and loading nccmath before mathtools to make it work).– Bernard
2 days ago
there's still some code of f(0) missing aha. its where I group f(1) ... f(4) as f(1)[1+a+...+a^4] etc. Also, I don't mind if it runs over two pages. I don't want the text to be squashed, all I wanted is to make use of the empty white space :)
– Maths
2 days ago
there's still some code of f(0) missing aha. its where I group f(1) ... f(4) as f(1)[1+a+...+a^4] etc. Also, I don't mind if it runs over two pages. I don't want the text to be squashed, all I wanted is to make use of the empty white space :)
– Maths
2 days ago
Refer to my code in the question, you'll see the part you missed :)
– Maths
2 days ago
Refer to my code in the question, you'll see the part you missed :)
– Maths
2 days ago
|
show 5 more comments
4
You should avoid \ on the last line of alignments. Perhaps the following is closer to what you want:
documentclass[11pt, a4paper]report
usepackageamsmath,array
begindocument
newcommandiuimkern1mu
beginequation*
setlengthextrarowheight3pt
begintabular c c c c c
& $0$ & $1$ & $2$ & $3$ & $4$\
cline1-6
$chi_0$ & $1$ & $1$ & $1$ & $1$ & $1$\
$chi_1$ & $1$ & $a$ & $a^2$ & $a^3$ & $a^4$\
$chi_2$ & $1$ & $a^2$ & $a^4$ & $a$ & $a^3$\
$chi_3$ & $1$ & $a^3$ & $a$ & $a^4$ & $a^2$\
$chi_4$ & $1$ & $a^4$ & $a^3$ & $a^2$ & $a$\
endtabular
endequation*
with $a = expfrac2piiu5$ hence $a^5=1$ with $|G|=5$.
Applying the definition of Fourier transform from Definition~3.1.2 we
have:
beginalign*
hatf(chi_0) &=f(0)+f(1)+f(2)+f(3)+f(4),\
hatf(chi_1) &=f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4),\
hatf(chi_2) &=f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4),\
hatf(chi_3) &=f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4),\
hatf(chi_4) &=f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4).
endalign*
Using Definition~3.1.3 we can compute the inverse Fourier transform
$f(t)$:
beginalign*
f(0)
&=frac15[ hatf(chi_0) + hatf(chi_1) + hatf(chi_2) +
hatf(chi_3) + hatf(chi_4)]\
&=frac15[f(0)+f(1)+f(2)+f(3)+f(4)]\
&qquad + frac15[f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4)]\
&qquad + frac15[f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4)]\
&qquad + frac15[f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4)]\
&qquad + frac15[f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)]
\
&= f(0)\
&qquad + fracf(1)5[1+a+a^2+a^3+a^4]\
&qquad +fracf(2)5[1+a+a^2+a^3+a^4]\
&qquad +fracf(3)5[1+a+a^2+a^3+a^4]\
&qquad +fracf(4)5[1+a+a^2+a^3+a^4]\
&=f(0).
endalign*
Similarly
beginalign*
f(1)
&= frac15Bigl[hatf(chi_0) + frac1ahatf(chi_1) +
frac1a^2hatf(chi_2) + frac1a^3hatf(chi_3) +
frac1a^4hatf(chi_4)Bigr]\
&=f(1),\
f(2)
&= frac15[hatf(chi_0) + a^2hatf(chi_1) +
a^4hatf(chi_2) + ahatf(chi_3) + a^3hatf(chi_4)]\
&=f(2), \
f(3)
&= frac15[hatf(chi_0) + a^3hatf(chi_1) +
ahatf(chi_2) + a^4hatf(chi_3) + a^2hatf(chi_4)]\
&=f(3),\
f(4)
&= frac15[hatf(chi_0) + a^4hatf(chi_1) +
a^3hatf(chi_2) + a^2hatf(chi_3) + ahatf(chi_4)]\
& =f(4).
endalign*
enddocument
answered May 1 at 15:09
Andrew SwannAndrew Swann
78.8k9138336
why did you push f(0) outwards? it wasn't necessary. but thanks for your solution
– Maths
May 1 at 15:42
1
The+'s should not be under the=, but to the right of it as they belong to that side of the equation. Whether you want to indent byqquadas I did, or the smallerquadis a matter of taste.
– Andrew Swann
May 1 at 18:28
add a comment |
4
You should avoid \ on the last line of alignments. Perhaps the following is closer to what you want:
documentclass[11pt, a4paper]report
usepackageamsmath,array
begindocument
newcommandiuimkern1mu
beginequation*
setlengthextrarowheight3pt
begintabular c c c c c
& $0$ & $1$ & $2$ & $3$ & $4$\
cline1-6
$chi_0$ & $1$ & $1$ & $1$ & $1$ & $1$\
$chi_1$ & $1$ & $a$ & $a^2$ & $a^3$ & $a^4$\
$chi_2$ & $1$ & $a^2$ & $a^4$ & $a$ & $a^3$\
$chi_3$ & $1$ & $a^3$ & $a$ & $a^4$ & $a^2$\
$chi_4$ & $1$ & $a^4$ & $a^3$ & $a^2$ & $a$\
endtabular
endequation*
with $a = expfrac2piiu5$ hence $a^5=1$ with $|G|=5$.
Applying the definition of Fourier transform from Definition~3.1.2 we
have:
beginalign*
hatf(chi_0) &=f(0)+f(1)+f(2)+f(3)+f(4),\
hatf(chi_1) &=f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4),\
hatf(chi_2) &=f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4),\
hatf(chi_3) &=f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4),\
hatf(chi_4) &=f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4).
endalign*
Using Definition~3.1.3 we can compute the inverse Fourier transform
$f(t)$:
beginalign*
f(0)
&=frac15[ hatf(chi_0) + hatf(chi_1) + hatf(chi_2) +
hatf(chi_3) + hatf(chi_4)]\
&=frac15[f(0)+f(1)+f(2)+f(3)+f(4)]\
&qquad + frac15[f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4)]\
&qquad + frac15[f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4)]\
&qquad + frac15[f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4)]\
&qquad + frac15[f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)]
\
&= f(0)\
&qquad + fracf(1)5[1+a+a^2+a^3+a^4]\
&qquad +fracf(2)5[1+a+a^2+a^3+a^4]\
&qquad +fracf(3)5[1+a+a^2+a^3+a^4]\
&qquad +fracf(4)5[1+a+a^2+a^3+a^4]\
&=f(0).
endalign*
Similarly
beginalign*
f(1)
&= frac15Bigl[hatf(chi_0) + frac1ahatf(chi_1) +
frac1a^2hatf(chi_2) + frac1a^3hatf(chi_3) +
frac1a^4hatf(chi_4)Bigr]\
&=f(1),\
f(2)
&= frac15[hatf(chi_0) + a^2hatf(chi_1) +
a^4hatf(chi_2) + ahatf(chi_3) + a^3hatf(chi_4)]\
&=f(2), \
f(3)
&= frac15[hatf(chi_0) + a^3hatf(chi_1) +
ahatf(chi_2) + a^4hatf(chi_3) + a^2hatf(chi_4)]\
&=f(3),\
f(4)
&= frac15[hatf(chi_0) + a^4hatf(chi_1) +
a^3hatf(chi_2) + a^2hatf(chi_3) + ahatf(chi_4)]\
& =f(4).
endalign*
enddocument
answered May 1 at 15:09
Andrew SwannAndrew Swann
78.8k9138336
why did you push f(0) outwards? it wasn't necessary. but thanks for your solution
– Maths
May 1 at 15:42
1
The+'s should not be under the=, but to the right of it as they belong to that side of the equation. Whether you want to indent byqquadas I did, or the smallerquadis a matter of taste.
– Andrew Swann
May 1 at 18:28
add a comment |
4
4
4
You should avoid \ on the last line of alignments. Perhaps the following is closer to what you want:
documentclass[11pt, a4paper]report
usepackageamsmath,array
begindocument
newcommandiuimkern1mu
beginequation*
setlengthextrarowheight3pt
begintabular c c c c c
& $0$ & $1$ & $2$ & $3$ & $4$\
cline1-6
$chi_0$ & $1$ & $1$ & $1$ & $1$ & $1$\
$chi_1$ & $1$ & $a$ & $a^2$ & $a^3$ & $a^4$\
$chi_2$ & $1$ & $a^2$ & $a^4$ & $a$ & $a^3$\
$chi_3$ & $1$ & $a^3$ & $a$ & $a^4$ & $a^2$\
$chi_4$ & $1$ & $a^4$ & $a^3$ & $a^2$ & $a$\
endtabular
endequation*
with $a = expfrac2piiu5$ hence $a^5=1$ with $|G|=5$.
Applying the definition of Fourier transform from Definition~3.1.2 we
have:
beginalign*
hatf(chi_0) &=f(0)+f(1)+f(2)+f(3)+f(4),\
hatf(chi_1) &=f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4),\
hatf(chi_2) &=f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4),\
hatf(chi_3) &=f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4),\
hatf(chi_4) &=f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4).
endalign*
Using Definition~3.1.3 we can compute the inverse Fourier transform
$f(t)$:
beginalign*
f(0)
&=frac15[ hatf(chi_0) + hatf(chi_1) + hatf(chi_2) +
hatf(chi_3) + hatf(chi_4)]\
&=frac15[f(0)+f(1)+f(2)+f(3)+f(4)]\
&qquad + frac15[f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4)]\
&qquad + frac15[f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4)]\
&qquad + frac15[f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4)]\
&qquad + frac15[f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)]
\
&= f(0)\
&qquad + fracf(1)5[1+a+a^2+a^3+a^4]\
&qquad +fracf(2)5[1+a+a^2+a^3+a^4]\
&qquad +fracf(3)5[1+a+a^2+a^3+a^4]\
&qquad +fracf(4)5[1+a+a^2+a^3+a^4]\
&=f(0).
endalign*
Similarly
beginalign*
f(1)
&= frac15Bigl[hatf(chi_0) + frac1ahatf(chi_1) +
frac1a^2hatf(chi_2) + frac1a^3hatf(chi_3) +
frac1a^4hatf(chi_4)Bigr]\
&=f(1),\
f(2)
&= frac15[hatf(chi_0) + a^2hatf(chi_1) +
a^4hatf(chi_2) + ahatf(chi_3) + a^3hatf(chi_4)]\
&=f(2), \
f(3)
&= frac15[hatf(chi_0) + a^3hatf(chi_1) +
ahatf(chi_2) + a^4hatf(chi_3) + a^2hatf(chi_4)]\
&=f(3),\
f(4)
&= frac15[hatf(chi_0) + a^4hatf(chi_1) +
a^3hatf(chi_2) + a^2hatf(chi_3) + ahatf(chi_4)]\
& =f(4).
endalign*
enddocument
answered May 1 at 15:09
Andrew SwannAndrew Swann
78.8k9138336
You should avoid \ on the last line of alignments. Perhaps the following is closer to what you want:
documentclass[11pt, a4paper]report
usepackageamsmath,array
begindocument
newcommandiuimkern1mu
beginequation*
setlengthextrarowheight3pt
begintabular c c c c c
& $0$ & $1$ & $2$ & $3$ & $4$\
cline1-6
$chi_0$ & $1$ & $1$ & $1$ & $1$ & $1$\
$chi_1$ & $1$ & $a$ & $a^2$ & $a^3$ & $a^4$\
$chi_2$ & $1$ & $a^2$ & $a^4$ & $a$ & $a^3$\
$chi_3$ & $1$ & $a^3$ & $a$ & $a^4$ & $a^2$\
$chi_4$ & $1$ & $a^4$ & $a^3$ & $a^2$ & $a$\
endtabular
endequation*
with $a = expfrac2piiu5$ hence $a^5=1$ with $|G|=5$.
Applying the definition of Fourier transform from Definition~3.1.2 we
have:
beginalign*
hatf(chi_0) &=f(0)+f(1)+f(2)+f(3)+f(4),\
hatf(chi_1) &=f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4),\
hatf(chi_2) &=f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4),\
hatf(chi_3) &=f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4),\
hatf(chi_4) &=f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4).
endalign*
Using Definition~3.1.3 we can compute the inverse Fourier transform
$f(t)$:
beginalign*
f(0)
&=frac15[ hatf(chi_0) + hatf(chi_1) + hatf(chi_2) +
hatf(chi_3) + hatf(chi_4)]\
&=frac15[f(0)+f(1)+f(2)+f(3)+f(4)]\
&qquad + frac15[f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4)]\
&qquad + frac15[f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4)]\
&qquad + frac15[f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4)]\
&qquad + frac15[f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)]
\
&= f(0)\
&qquad + fracf(1)5[1+a+a^2+a^3+a^4]\
&qquad +fracf(2)5[1+a+a^2+a^3+a^4]\
&qquad +fracf(3)5[1+a+a^2+a^3+a^4]\
&qquad +fracf(4)5[1+a+a^2+a^3+a^4]\
&=f(0).
endalign*
Similarly
beginalign*
f(1)
&= frac15Bigl[hatf(chi_0) + frac1ahatf(chi_1) +
frac1a^2hatf(chi_2) + frac1a^3hatf(chi_3) +
frac1a^4hatf(chi_4)Bigr]\
&=f(1),\
f(2)
&= frac15[hatf(chi_0) + a^2hatf(chi_1) +
a^4hatf(chi_2) + ahatf(chi_3) + a^3hatf(chi_4)]\
&=f(2), \
f(3)
&= frac15[hatf(chi_0) + a^3hatf(chi_1) +
ahatf(chi_2) + a^4hatf(chi_3) + a^2hatf(chi_4)]\
&=f(3),\
f(4)
&= frac15[hatf(chi_0) + a^4hatf(chi_1) +
a^3hatf(chi_2) + a^2hatf(chi_3) + ahatf(chi_4)]\
& =f(4).
endalign*
enddocument
answered May 1 at 15:09
Andrew SwannAndrew Swann
78.8k9138336
edited May 1 at 18:25
answered May 1 at 15:09
Andrew SwannAndrew Swann
78.8k9138336
answered May 1 at 15:09
Andrew SwannAndrew Swann
78.8k9138336
answered May 1 at 15:09
Andrew SwannAndrew Swann
78.8k9138336
78.8k9138336
why did you push f(0) outwards? it wasn't necessary. but thanks for your solution
– Maths
May 1 at 15:42
1
The+'s should not be under the=, but to the right of it as they belong to that side of the equation. Whether you want to indent byqquadas I did, or the smallerquadis a matter of taste.
– Andrew Swann
May 1 at 18:28
add a comment |
why did you push f(0) outwards? it wasn't necessary. but thanks for your solution
– Maths
May 1 at 15:42
1
The+'s should not be under the=, but to the right of it as they belong to that side of the equation. Whether you want to indent byqquadas I did, or the smallerquadis a matter of taste.
– Andrew Swann
May 1 at 18:28
why did you push f(0) outwards? it wasn't necessary. but thanks for your solution
– Maths
May 1 at 15:42
why did you push f(0) outwards? it wasn't necessary. but thanks for your solution
– Maths
May 1 at 15:42
1
1
The
+'s should not be under the =, but to the right of it as they belong to that side of the equation. Whether you want to indent by qquad as I did, or the smaller quad is a matter of taste.– Andrew Swann
May 1 at 18:28
The
+'s should not be under the =, but to the right of it as they belong to that side of the equation. Whether you want to indent by qquad as I did, or the smaller quad is a matter of taste.– Andrew Swann
May 1 at 18:28
add a comment |
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Unrelated to the issue, but switching from tabular to array you can remove all the repeated $ signs.
– leandriis
May 1 at 16:37