[SOLVED] IT代考 Lecture 15 :

Lecture 15 :
Orthogonal Expansions
* This material is from section 3.2
I . Orthogonal Functions

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A. M otivation From PDE
* Recall from last time we introduced the notion
of Solving Ut=Khxx , 0 < x < IT , Ula,t)= UCH,t)=ovia seriesofthe form UH,t)=-2,1ane-kñtsinlnx)which lead us toconsideringn>(1 expansions
Ula,a)= No(a)= 2T ansinlnx) N>/ I
* The key step to computing an is to use the formulas s.mn>c)sum,ydx =)”% , n=m
so that an__ F)ITsincnxuolxdx.

B. Orthogonal Functions
Defy: let IT be a read vector space . An ” inner product” on is a rule 4;-):xxx- →IR suchthat
i) Lf,f>7,0 all fat, and Lf,f>=o off f=o. ii) Lf, g)=Lg.f) all figC-4 .
Iii) Laftbg,h> = a 4th) 1- big,h> all f.g.HEY , a,btlR.
ez: The dot
product on 11=112″ is an inner product. figi .
eg_: let Atmnxn , AT=A and all eigenvalues R of
A satisfy ×> u . Then Lf,g> = fTAg
all f,gtlR^ is an inner product
lety={f:[a.b)→IR /fis continuous}
Then Lf,g>= Sabfydx B- an inner product.
⇒ This is called the ” LZ inner
Note : In this case V is infinite dimension-1 .
e.g. fn=snlnx) are all orthogonal for [a.b)= [0,1T].

Deff: let 11 be a vector space with an inner product L , > . We say a collection of vectors
fn are – if Lfn ,fm> = o all n -4M orthogonal .
ed: The standard basis ej=/ Ith slot on Ipn
orthogonal .
Ed: fn= smcnx) are LZ orthogonal on [0,1T].
eg-: gn= cos (n a) are i
orthogonal o n [ 0,1T]
ez fn ,gn a s defined above a re not orthogonal
on [0,1T]. For example SIS.nl?i)cosl2x)dx
use : SintA)cos /B) =I(SinlAtB)tsinlA- B)) .
= { SF (Sinisi) – sink)) dx
=L ( – } costs>c) lot + cost>DIE)

ed : the collection
{fn , gn} above is LZ orthogonal Note that Sirena>Cosima) is
so I sihlnn> cosima)dx=o .
always odd ,
Also Sink>c)sinlmx) is ever so
Sttsinlnnsinlmxsdx
25’T )dx= /IT osinlnxsinlm.sc Lo
s i m i l a r l y { C o s t n e r) , c o s ( m a) > = f I T , n = m ) Lo, n#m .
Orthogolal
expansions
Defy : A collection o f orthogonal vectors {fn} is called ” complete” if Lfn,f>=o for all n implies f=o.
* If {fn} is compute then we hope to be
able to expand general f= anfn choosing an= . In general this
is delicate because w e have a n infinite sum

in a n infinite dimensional
*Defy: Let V={f:[a.b)→IR I fcontinuous}
let {fn} be an orthogonal collection (e.g. sirloin)) .
,(a)=I’fn→f-pointingif n=1
F-×, (a)= f-(a)- ¥1anfnlx) → o n=1
all set (a)b) .
We say Sx, (a)→ Fbi] uniformly if sup 1Each/→ 0 KEG, b)
as 111→v0 .
we Sx→f in the L” or “leastsquares” say, —
sense if> Fda → o
* It turns out the last notion (least-squares) is
I t works fo r arbitrary continuous while the pointwise notion can break down
the most robust.
at a desc set of points.

Some inequalities involving Fourier Coefficients
A. Least squares
approximations
* let {fn} be a n orthogonal collection , and
It “‘”) set an= . set Lf,f>=/If11 (L norm .
Then 11 Em 112–11 f- Éanfnhz . let {bn} be
orther collection of
coefficients . Ther
Ilf – ¥ bnfn112=111-112-2-2 bn Lfn ,f> + 2- BE < fn.fm>
=/If112 anbncfn.fm> 2T bil-n.fi
= 111-112 –
2T (an-bnYL-fn.fr> n=,
=/If – If anfn112 t 2T (an- bn) (fn,fn> . n=, n=/
* This shows three things:

1) 11 f- Eianfnll £11 f- 2T bnfnll {bn}
n=I n=,any, N
11 f-I.bnfnll is mEed by taking an n=,
orthonormal (om) then
aicfn.fm> f 111-112 ( Bessel’s Inequality) (take bn=o toprove it).
particular if ,
i.e. {fn} are
Lfn,fn>=I ,
* Note: that this is a weaker condition than ¥ /ant Lou , ,
so convergence is tricky even when /fnl EC uniformly.
an ON then 11 -20 3) INf {fn} is i fEmll f
Éai IFIP because 11EμÑ=111-112 I’ ai
It can be shown {fn} is complete iff É’ ai-41ft . n=,

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