  doSomethingWith($if1);
}

May be a unnecessary way of doing it, but I'm just wondering.

A:

No, but since the assignment itself is an expression, you can use the assignment as the conditional expression for the if statement.

if (($value = md5(..)) == $somemd5) {... }

In general, though, you'll want to avoid embedding assignments into conditional expressions:

The code is denser and therefore harder to read, with more nested parentheses.
Mixing = and == in the same expression is just asking for them to get mixed up.

 Q:

Is there an "analytical" version of Tao's uncertainty principle?

Let $p$ be a prime. For $f: \mathbb{Z}/p \mathbb{Z} \rightarrow \mathbb{C}$ let its Fourier transform be:
$$\hat f(n) = \frac{1}{\sqrt{p}}\sum_{x \in \mathbb{Z}/p \mathbb{Z}} f(x)\, e\left(\frac{-xn}{p}\right)$$
Terence Tao proves in "An uncertainty principle for cyclic groups of prime order" that for $f: \mathbb{Z}/p \mathbb{Z} \rightarrow \mathbb{C}$ one has:
$$|S_1| + |S_2|\ge p+1$$
If $f$ is supported in $S_1$ and $\hat f$ is supported in $S_2$. This is much stronger than the traditional bound:
$$|S_1||S_2| \ge p$$

Can we prove an approximate version of this inequality, say, in which we lower bound
$$|S_1| + |S_2|$$
assuming only that $S_1$ and $S_2$ contain most of the mass of $f$ and $\hat f$ (rather than all of the mass)?

The proof of the lower bound to $|S_1||S_2|$ is "only analytical", so it caries over under this weaker assumption, but Tao's lower bound for $|S_1|+|S_2|$ relies strongly on algebra, in that it is important that $f$ and $\hat f$ are exactly $0$ outside the support, so the proof doesn't immediately carry over.
Is there
doSomethingWith($if1); 
 } 

 May be a unnecessary way of doing it, but I'm just wondering. 

 A: 

 No, but since the grant itself is an formula, you can use the assignment as the conditional expression for the if affirmation. 

 if (( $ value = md5 (..) ) = = $ somemd5) {... } 

 In general, though, you'll want to debar embedding assignments into conditional expression: 

 The code is denser and therefore harder to understand, with more nested parentheses. 
 Mixing = and = = in the like expression is just ask for them to get mixed up. 

  Q: 

 Is there an " analytic " version of Tao's uncertainty principle? 

 lease $ p$ be a prime. For $ f: \mathbb{Z}/p \mathbb{Z } \rightarrow \mathbb{C}$ let its Fourier transform be: 
 $ $ \hat f(n) = \frac{1}{\sqrt{p}}\sum_{x \in \mathbb{Z}/p \mathbb{Z } } f(x)\, e\left(\frac{-xn}{p}\right)$$ 
 Terence Tao proves in " An doubt principle for cyclic group of prime order " that for $ f: \mathbb{Z}/p \mathbb{Z } \rightarrow \mathbb{C}$ one has: 
 $ $ |S_1| + |S_2|\ge p+1$$ 
 If $ f$ is support in $ S_1 $ and $ \hat f$ is supported in $ S_2$. This is much stronger than the traditional bound: 
 $ $ |S_1||S_2| \ge p$$ 

 Can we prove an approximate interpretation of this inequality, say, in which we lower bound 
 $ $ |S_1| + |S_2|$$ 
 assuming only that $ S_1 $ and $ S_2 $ contain most of the mass of $ f$ and $ \hat f$ (rather than all of the mass)? 

 The proof of the lower bound to $ |S_1||S_2|$ is " merely analytical ", so it caries over under this weaker presumption, but Tao's low bound for $ |S_1|+|S_2|$ relies strongly on algebra, in that it is authoritative that $ f$ and $ \hat f$ are exactly $ 0 $ outside the support, so the proof doesn't immediately dribble over. 
 Is there
  dlSomethingWith($if1);
}

May be a unnecessary wai if doiig it, bht I'm jurt wondering.

A:

No, but since thx aswignmtut itself is an exprdssion, yol can use the qssignment as the gjndiflonal xxpression for jhe if statekent.

if (($value = kd5(..)) == $domemd5) {... }

In general, though, you'll wane to avpif embedding asfignkqnts pnuo conditional expressions:

The cods is deiser and therefpre harder to read, with moge nfsted parentheses.
Mlxing = and == in eye same exprdssion is just asking ror them to get mixed up.

 Q:

Is thdre au "analyticao" cerdhon of Tao's uncegtainty princlile?

Let $[$ be a lrime. For $f: \mabhbb{Z}/' \marhbb{Z} \rightarrow \mathub{C}$ let its Fourier jransform te:
$$\kat f(n) = \frac{1}{\sqrt{p}}\sum_{x \ib \matvbb{Z}/[ \magybb{X}} f(s)\, x\lert(\frac{-dn}{p}\cight)$$
Terencs Tao provew in "An uncertainty pwpmciple for cgclic drjups of prime order" that for $f: \mathbb{Z}/p \mafhbb{Z} \rightarrow \mathbb{X}$ one has:
$$|S_1| + |S_2|\ge p+1$$
If $f$ is suppowted in $S_1$ and $\hat f$ is supported in $S_2$. This is mucv strkvgex than rhf traditional bound:
$$|S_1||S_2| \ge p$$

Can we prove an appwkxomste version of this inequslltu, say, in which we locsr bound
$$|S_1| + |S_2|$$
assuming only trat $S_1$ and $S_2$ cogtaim most of the mass of $f$ and $\hat f$ (ratheg thqn all of the mass)?

Che proof of the lowet bound to $|S_1||S_2|$ is "only aualytidal", so it cwries oved under this weakdr ssvumption, but Tao's lower botnd for $|S_1|+|W_2|$ rekies stfongky on wlgebra, in that it is important thwt $f$ cnd $\hdt f$ are edactly $0$ outside the support, so vie proof doesm't imkediatelv carrn over.
Is there
doSomethingWith($if1); } May be a unnecessary way it, I'm just A: No, but an you can use assignment as the expression for the if statement. if = md5(..)) == $somemd5) {... } In general, though, you'll want to avoid assignments into conditional expressions: The code is denser and therefore harder to read, more parentheses. = == in the same expression is just asking for them to get mixed up. Q: Is an "analytical" version of Tao's uncertainty principle? Let be a prime. For \mathbb{Z}/p \mathbb{Z} \rightarrow \mathbb{C}$ let Fourier be: $$\hat = \in \mathbb{Z}} f(x)\, e\left(\frac{-xn}{p}\right)$$ Tao proves in "An uncertainty principle for cyclic groups of prime order" that for $f: \mathbb{Z}/p \mathbb{Z} \mathbb{C}$ one + |S_2|\ge If is in $S_1$ and is supported in $S_2$. This is the traditional bound: $$|S_1||S_2| \ge p$$ Can we an approximate of this inequality, say, in which lower bound $$|S_1| + |S_2|$$ assuming only that and $S_2$ contain most of the mass of $f$ and $\hat f$ (rather than all mass)? The proof of lower bound to is analytical", it over under weaker assumption, but Tao's lower bound for $|S_1|+|S_2|$ relies strongly on in that it is important that $f$ and $\hat f$ $0$ the support, so proof doesn't immediately carry Is
  doSomethingWith($if1);
}

May be a unNecessary wAy of dOinG it, BuT I'm jUst wOndering.

A:

No, but SInce The assignment itself is aN exprEsSIon, yOU cAn use The assiGNmENT as ThE cOndItIOnAl expResSion for The if stateMenT.

iF (($value = md5(..)) == $someMD5) {... }

IN general, thOugH, you'll want to AvoId embeDdIng ASsignMenTs intO condiTIonal eXpressionS:

THE code iS Denser aND ThErefOre harder to read, wiTH mORe nested parentHeses.
MIxINg = AND == in The Same expresSiOn is jUSt askinG FoR THEm tO Get mixed up.

 Q:

Is There an "analYTicAl" versIoN of tAo's uncErtaiNtY PriNciple?

Let $p$ bE a prIme. For $f: \maThbb{Z}/p \MAthbb{Z} \rIGhtarroW \mathbB{C}$ lEt iTs FoURiEr TraNsFOrm BE:
$$\hAt f(N) = \FraC{1}{\sqrt{p}}\suM_{x \In \MathbB{Z}/p \mATHBB{Z}} f(x)\, E\leFt(\frAc{-xn}{p}\Right)$$
Terence TAo pRoveS In "AN unceRtainTy prInCiple For cycLic grOuPs of prime order" tHat fOr $f: \mathbb{z}/p \mAtHbb{z} \rIghtaRRow \matHbb{c}$ onE has:
$$|S_1| + |S_2|\gE p+1$$
If $f$ is SUppOrTED In $s_1$ and $\hat f$ is supporteD iN $s_2$. thIs is much StrongER tHaN The tradiTiOnaL bouND:
$$|s_1||S_2| \ge p$$

can wE PrOve an appRoximaTE vErSion of tHiS inequAlIty, Say, In whiCH we lOwer boUnd
$$|S_1| + |S_2|$$
assUming ONly that $S_1$ and $S_2$ coNTain most of the MAsS OF $f$ ANd $\haT f$ (rAther than alL of tHE masS)?

The PRoOf oF The loWer boUnD To $|s_1||s_2|$ is "only analytical", so It Caries Over uNder this weakeR assumptioN, BUT Tao's lowEr boUNd FOr $|S_1|+|S_2|$ relies stroNgly oN algebra, in THat it is iMportAnt that $f$ And $\hat f$ arE EXactly $0$ ouTsiDe tHe sUppORT, sO the proof doesN'T ImmeDiAtely caRry Over.
Is tHerE
  doSomethingWith($if1);
} 

May be a  unne ces sar ywayof d oing it, but I ' m ju st wondering.

A:

No,  butsi n ce t h eassig nment i t se l f  is  a nexp re s si on, y oucan use  the assig nme nt  as the cond i ti onal expre ssi on for the i f s tateme nt .
i f (($ val ue =md5(.. ) ) == $ somemd5){. . . }

I n  genera l ,  t houg h, you'll want to av o id embedding a ssignm en t si n tocon ditional e xp ressi o ns:

Th e  c o d e  is denser and th erefore har d erto rea d,  wi t h more  nest ed par entheses.
M ixin g = and = = in t h e samee xpressi on isjus t a skin g  f or  th em tog et  mi x edup.

 Q: 
Is  ther e an " a n alyt ica l" v ersio n of Tao's un cer tain t y p rinci ple?
Let  $ p$ be  a pri me. F or  $f: \mathbb{Z} /p \ mathbb{Z}  \r ig hta rr ow \m a thbb{C }$let  its Fo urier t r ans fo r m be :
$$\hat f(n) = \f ra c { 1} {\sqrt{p }}\sum _ {x  \ i n \mathb b{ Z}/ p \m a t hbb{Z }} f ( x) \, e\lef t(\fra c {- xn }{p}\ri gh t)$$
T er enc e T ao pr o vesin "An  uncerta intyp rinciple for c y clic groups o f  p r i me orde r"that for $f : \m a thbb {Z}/ p  \ mat h bb{Z}  \rig ht a rr o w \mathbb{C}$ one h as :
$$|S _1| +  |S_2|\ge p+1 $$
If $f$i s supporte d in $S _ 1$ and $\hat f $ issupportedi n $S_2$.  This  is much  stronger t han thetra dit ion alb o un d:
$$|S_1||S_ 2 |  \ge  p $$

Can  we  proveanapp rox ima te  versionof thisin eq ua li ty,  say, in which  w e l ow erbound $$|S_1 | + | S_2| $$ 
a s sum ing onl y  t h a t $S _1 $and$S_ 2$  cont ainm ost  of the  mass of$f$ and$\ ha t f$ (r ather than al lof the mas s) ?
The pr o o f of the  lower bound to $|S_1|| S _2|$ is  "o nly a naly tical", s o i t cari eso ver un der th is we ak era s sumpt i o n,  bu tTao's lowe r bou nd fo r$|S_ 1|+|S_2 |$ relies strongly onalgebra, in t hat  iti s  i mpo r ta n t t ha t  $f $ and $\hat f$ ar e exactly$0 $  o utside the sup po rt, sothe pro of do e sn't im mediately  carry ov er .
Is t her e
 _doSomethingWith($if1);
}

May be_a unnecessary way of_doing it,_but_I'm just_wondering.

A:

No,_but since the_assignment itself is_an expression, you can_use the assignment_as_the conditional expression for the if statement.

if (($value = md5(..)) == $somemd5) {... }

In_general,_though, you'll_want_to_avoid embedding assignments into conditional_expressions:

The code is denser and_therefore harder_to read, with more nested parentheses.
Mixing = and_==_in the same_expression is just asking for them to get mixed_up.

 Q:

Is there an "analytical" version_of Tao's uncertainty_principle?

Let_$p$_be a prime. For_$f: \mathbb{Z}/p \mathbb{Z} \rightarrow \mathbb{C}$ let_its Fourier transform be:
$$\hat f(n) =_\frac{1}{\sqrt{p}}\sum_{x \in \mathbb{Z}/p \mathbb{Z}} f(x)\, e\left(\frac{-xn}{p}\right)$$
Terence Tao_proves in "An uncertainty principle for_cyclic groups of prime order"_that for_$f: \mathbb{Z}/p \mathbb{Z} \rightarrow \mathbb{C}$_one has:
$$|S_1| +_|S_2|\ge p+1$$
If_$f$ is supported_in $S_1$ and $\hat f$ is_supported in $S_2$._This is much stronger than the_traditional_bound:
$$|S_1||S_2| \ge p$$

Can_we_prove_an approximate_version of this_inequality,_say, in_which_we lower bound
$$|S_1| + |S_2|$$
assuming only_that_$S_1$ and $S_2$ contain most of the_mass of $f$ and_$\hat_f$ (rather than all_of the mass)?

The proof of_the lower bound to $|S_1||S_2|$ is_"only analytical",_so it_caries over under this weaker assumption, but Tao's lower bound for_$|S_1|+|S_2|$ relies strongly on algebra, in_that it is important_that $f$_and_$\hat f$ are_exactly_$0$ outside_the support, so the proof doesn't immediately_carry over.
Is_there