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Linear Algebra, Theory And Applications, 2012a

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182 SPECTRAL THEORY<br />

and so if the new variables are given by<br />

⎛<br />

⎜<br />

⎝<br />

⎞<br />

− √ 1 √6 1 √3 1<br />

2<br />

2<br />

0 √6 − √ 1 ⎟<br />

3 ⎠<br />

√1<br />

√6 1 √3 1<br />

2<br />

⎛<br />

⎝<br />

x ′<br />

y ′<br />

z ′<br />

⎞ ⎛<br />

⎠ = ⎝<br />

it follows that in terms of the new variables the quadratic form is 2 (x ′ ) 2 − 4(y ′ ) 2 +8(z ′ ) 2 .<br />

You can work other examples the same way.<br />

7.7 Second Derivative Test<br />

Under certain conditions the mixed partial derivatives will always be equal. This astonishing<br />

fact was first observed by Euler around 1734. It is also called Clairaut’s theorem.<br />

Theorem 7.7.1 Suppose f : U ⊆ F 2 → R where U is an open set on which f x ,f y ,f xy and<br />

f yx exist. Then if f xy and f yx are continuous at the point (x, y) ∈ U, itfollows<br />

f xy (x, y) =f yx (x, y) .<br />

Proof: Since U is open, there exists r>0 such that B ((x, y) ,r) ⊆ U. Now let |t| , |s| <<br />

r/2,t,s real numbers and consider<br />

h(t)<br />

Δ(s, t) ≡ 1 { }} { { }} {<br />

st { f (x + t, y + s) − f (x + t, y) − (f (x, y + s) − f (x, y))}. (7.16)<br />

Note that (x + t, y + s) ∈ U because<br />

x<br />

y<br />

z<br />

⎞<br />

⎠ ,<br />

h(0)<br />

|(x + t, y + s) − (x, y)| = |(t, s)| = ( t 2 + s 2) 1/2<br />

≤<br />

( r<br />

2<br />

4 + r2<br />

4<br />

) 1/2<br />

= r √<br />

2<br />

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