Spivak's proof of Cauchy Schwarz

  • #1
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I was browsing through Spivak's Calculus book and found in a problem a very simple way to prove the cauchy schwarz inequality.

Basically he tells to substitute x=xᵢ/[√(x₁²+x₂²)] and similarly for y (i=1 and 2), put into x^2 + y^2 >= 2xy. Add the two cases and we get the result.

The problem is 18 C of page 18 of prologue part of the book. The book can be found in archive.org.

<Moderator's note: Link removed due to copyright violation.>

Here is my proof

https://ibb.co/dc3rUJ


Can any one tell me from where does Spivak get this substitution x=xᵢ/[√(x₁²+x₂²)] and similar one for y ?

Thanks.Sorry no LaTex. If you don't understand my handwriting please ask. Links will expire after 3 days.

Any help is appreciated.
 
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Answers and Replies

  • #2
mathwonk
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in vector notation, where X = (x1,x2) and Y = (y1,y2,) the inequality states that |X.Y| ≤ |X||Y|, where X.Y is the dot product of the vectors X and Y, i.e. x1y1+x2y2, and |X| and |Y| are the lengths of the vectors, i.e. |X| = sqrt(X.X) and |Y| = sqrt(Y.Y). now by dividing through by the lengths you can see that it suffices to prove it for vectors of length 1, i.e. assuming X and Y have length one, we want to show that |X.Y| ≤ 1. but applying the given inequality we have x1y1 ≤ x1^2 + y1^2, and similarly for x2,y2. So we can just add and get 2(X.Y) on the left and x1^2 + x2^2 + y11^2 + y2^2 = 1+1 = 2, on the right. so he is reducing the proof to vectors of length one by dividing each component by the length of the vector.
 
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  • #3
mathwonk
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by the way, since dot products are bilinear and commutative, we always have |X-Y|^2 = (X-Y).(X-Y) = X^2 - 2X.Y + Y^2, and since |X-Y| is the length of the third side of the triangle with sides X and Y emanating from the origin, the law of cosines says that also |X-Y|^2 = |X|^2 + |Y^2| - 2|X||Y|cos(t) where t is the angle formed by X and Y at the origin. And since |cos(t))| ≤ 1, we are done. I.e. thus X.Y = |X||Y|cos(t) ≤ |X||Y|.

Moreover, the law of cosines (a generalization of the pythagorean theorem to arbitrary triangles) is proved in Euclid, Book II, Propositions 12 and 13, so the reason behind this fact greatly predates Cauchy and Schwartz.

I.e. euclid shows that the difference between |X-Y|^2 and |X|^2 + |Y|^2, is twice the area of a rectangle, one of whose sides is say |X| and the other is a side of a right triangle with hypotenuse |Y|. Since that side is shorter than the hypotenuse, we have that 2|X.Y| = |X|^2 + |Y|^2 - |X-Y|^2 ≤ 2|X||Y|.

I personally consider it a huge loss that plane geometry is no longer ordinarily taught from Euclid, by far the most informative book on the topic I have ever seen. Anyone wishing to study it is recommended to take Hartshorne's Geometry: Euclid and beyond, as a marvelous guide.
 
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  • #4
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I don't think you can use Euclidean geometry proofs in vector space analysis,
 
  • #5
mathwonk
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i would suggest reading the proof. or perhaps you would feel better if you first checked the euclidean axioms using vector methods. it is "well known" that R^2 is one of many models for euclidean geometry. I recommend reading Hartshorne, Geometry: Euclid and Beyond, for more insight on how euclidean geometry is related to coordinate geometry. But to be honest I had the same reservations you have a few years ago. We are taught math in such rigidly separated courses, we don't feel comfortable putting them together.

But fact: all the axioms of euclidean geometry, as enhanced by Hilbert, are satisfied by the lines and points and triangles of cartesian geometry in R^2, i.e. they are theorems in vector analysis, hence all proofs dependent only on the euclidean axioms are also true in vector geometry of R^2.

but this is technical. my point was, that even if stated in a different language, the fact at issue was known for over 2,000 years. if you wish, you may be able to see that euclid also anticipated newton's definition of a tangent line as a limit of secants, in prop.16 book 3.
 
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  • #6
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Thanks Mathwonk. Sorry for being late.

Never realized Spivak would use vector operations in algebraic expressions.

I thought that the form was derived from Lagrange's Identity, algebraic form.
 
  • #7
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I personally consider it a huge loss that plane geometry is no longer ordinarily taught from Euclid, by far the most informative book on the topic I have ever seen. Anyone wishing to study it is recommended to take Hartshorne's Geometry: Euclid and beyond, as a marvelous guide.

Right.

Our entire modern mathematics is based on Euclid's axiomatic approach, Euclid's axioms, his geometry. We don't remember him, but he is omnipresent in the most basic calculations.

Euclid marvelously derived the theorems without any tool that we are so acquainted with.

A school level course on geometry includes cursory proof of his theorems. Therefore we all pay, due to the system.

Geometry: Euclid and beyond is a classic, but never had the spare time to read it. And Elements(1-13) is a must,if not possible Hall/Stevens or Barnard/Child.


My respects to his unparalleled genius.
 

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