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单词 PolyaVinogradovInequality
释义

Pólya-Vinogradov inequality


Theorem 1.

For m,nN and p a positive odd rationalprime,

|t=mm+n(tp)|<plnp.
Proof.

Start with the following manipulations:

t=mm+n(tp)=1pt=0p-1x=mm+na=0p-1(tp)e2πia(x-t)/p=1pa=1p-1x=mm+ne2πiax/pt=0p-1(tp)e-2πiat/p

The expression t=0p-1(tp)e-2πiat/p is just a Gauss sumDlmfPlanetmath, and has magnitudep. Hence

|t=mm+n(tp)||ppa=1p-1x=mm+ne2πaix/p|=|ppa=1p-1e2πiam/px=0ne2πiax/p||ppa=1p-1e2πian/p-1e2πia/p-1|
=|ppa=1p-1eπian/psin(πan/p)eπia/psin(πa/p)|ppa=1p-1|1sin(πa/p)|ppa=1p-112a/p

Here x denotes the absolute valueMathworldPlanetmathPlanetmathPlanetmath of the differencebetween x and the closest integer to x, i.e. x=infz𝐙{|x-z|}.

Since p is odd, we have

12a=1p-11a/p=0<a<p2pa=pa=1p-121a

Now ln2x+12x-1>1x for x>1; to prove this, itsuffices to show that the functionMathworldPlanetmath f:[1,)𝐑given by f(x)=xln2x+12x-1is decreasing and approaches 1 as x. To prove thelatter statement, substitute v=1/x and take the limit as v0 using L’Hôpital’s rule. To prove the former statement,it will suffice to show that f is less than zero on the interval[1,). But f(x)0 as x andf is increasing on [1,), since f′′(x)=-44x2-1(1-4x2+14x2-1)>0 for x>1, so fis less than zero for x>1.

With this in hand, we have

|t=mm+n(tp)|pppa=1p-121a<pa=1p-12ln2a+12a-1=plnp.

References

  • 1 Vinogradov, I. M., Elements of Number TheoryMathworldPlanetmath, 5th rev. ed., Dover, 1954.
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