Test inequality
$ sec (x) ge 1+ (x ^ 2/2) $
Test inequality
$ sec (x) ge 1+ (x ^ 2/2) $
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Dice $$ x ^ 2u_ {xx} + 2xyu_ {xy} – 3y ^ 2u_ {yy} – 2xu_x + 4yu_y + 16x ^ 4u = 0 $$
I have to find the canonical form. The determinant is the following $$ D = 16x ^ 2y ^ 2 $$
I considered cases when $ x $ Y $ and $ can be $ 0 $. When calculating for the remaining case (where both $ x $ Y $ and $ It is not $ 0 $ and the type is hyperbolic), I'm not getting its canonical form. I'm receiving this
$$ (- 3y ^ 2 – 6x ^ 3y + 9x ^ 6) u _ { xi xi} + (-3y ^ 2 + frac {2y} {x} + frac {1} {x ^ 2} ) u _ {eta} + (-6y ^ 2 – 6x ^ 2 + frac {2y (1-3x ^ 4)} x) u _ { eta xi} + ( frac {-4} {x} + 4y) u_ eta + 4yu_ xi + 16x ^ 4u = 0 $$
But the point is that this does not meet the requirements for the canonical form of hyperbolic type.
Any help would be appreciated.
Leave $ x; Y; z in R ^ + $ such that $ x + y + z + 2 = xyz $. Such that $$ x + y + z + 6 ge 2 ( sqrt {xy} + sqrt {yz} + sqrt {xz}) $$
This inequality is not homogeneous and look at the condition that I thought
I would replace the variables $ x; Y; z $ such that
+)$ x ^ 2 + y ^ 2 + z ^ 2 + 2xyz = 1 $. Leave $ x = frac {2a} { sqrt { left (a + b right) left (a + c right)}} $
+)$ xy + yz + xz + xyz = 4 $. Leave $ a = frac {2 sqrt {xy}} { sqrt { left (y + z right) left (x + z right)}} $.Leave $ x = frac {2a} {b + c} $
but failed. Please explain to me how I can obtain this substitution (if I have a solution by substitution)
I also tried to solve it by $ u, v, w $.Leave $ sum_ {cyc} x = 3u; sum_ {cyc} xy; Pi_ {cyc} a = w ^ 3 (3u + 2 = w ^ 3; u, v, w> 0) $ so $ u le w ^ 3-3u $ or $ 4u le w ^ 3 $ but stagnant (I'm very bad at $ uvw $)
Leave $ X $ be a normed vector space and $ A $ be a subset of $ X $. $ conv (A) $
It is called the intersection of all convex subsets of $ X $ that contain
$ A $a) Show that conv (A) is a convex set
b) show that
$$ conv (A) = { sum_ {i = 1} ^ n lambda _i x_i: sum_ {i = 1} ^ n lambda_i = 1,
lambda_i ge 0, x_i in A, i = 1, cdots, n } $$c) yes $ A $ it is compact then it is $ conv (A) $ compact?
d) Show that if $ A subseteq mathbb {R} ^ n $ it is compact then $ conv (A) $ is
compact
a) Take two elements $ a $ Y $ b $ at the intersection of all convex subsets of $ X $ that contain $ A $. Now take $ lambda a + (1- lambda) b $. It is contained in each of the subsets of $ X $ that contain $ A $, therefore it is contained in the intersection. Q.E.D.
second)
$ leftarrow $:
Suppose by induction that $ sum_ {i = 1} ^ n lambda_i x_i $ It belongs to conv (A). We must prove that $ sum_ {i = 1} ^ {k + 1} lambda_i x_i $, with $ sum_ {i = 1} ^ {k + 1} = 1 $ It also belongs.
$$ sum_ {i = 1} ^ {k + 1} lambda_i x_i = sum_ {i = 1} ^ {k} lambda_i x_i + lambda_ {k + 1} x_ {k + 1} = sum_ {i = 1} ^ {k} lambda_i x_i + (1- sum_ {i = 1} ^ n lambda_i) x_ {k + 1} $$
Now choose $ delta $ such that $ delta sum_ {i = 1} ^ {k} lambda_i = 1 $, so
$$ frac { delta} { delta} left ( sum_ {i = 1} ^ {k} lambda_i x_i + (1- sum_ {i = 1} ^ n lambda_i) x_ {k + 1 } right = = left ( frac {1} { delta} sum_ {i = 1} ^ {k} ( delta lambda_i) x_i + frac {1} { delta} ( delta-1) x_ {k + 1} right) = \ frac {1} { delta} x + left (1- frac {1} { delta} right) x_ {k + 1} $$
which is a collection of elements of $ A $ that adds up to $ 1 $
$ rightarrow $:
I can only say that $ x in conv (A) $, so $ x = 1x + 0 cdot all $ Therefore, it is a combination that adds to $ 1 $ of elements of $ A $?
by do), give me a clue: show that conv (A union B) is the image through a continuous function of the compact $ {( alpha, beta; alpha, beta ge 0, alpha + beta = 1) } times A times B $. I do not understand how to show this.
re) Some clue?
Leave $ mathbb S_ {n-1} $ be the unitary sphere in $ mathbb R ^ n $ Y $ z_1, ldots, z_n $ be a sample i.i.d of $ mathcal N (0, 1) $.
Dice $ epsilon> 0 $ (It can be assumed that it is very small), which is reasonable upper limit for the tail probability $ P ( sup_ {a in mathbb S_ {n-1}} sum_ {i = 1} ^ na_i ^ 2z_i ^ 2 ge epsilon) $ ?
Using ideas from this other answer (MO link), you can set the not uniform Limit of anti-concentration: $ P ( sum_ {i = 1} ^ na_i ^ 2z_i ^ 2 le epsilon) le sqrt {e epsilon} $ for all $ a in mathbb S_ {n-1} $.
The uniform analog is another story. Can coverage numbers be used?
The book Theory of Numbers by G H Hardy, et al. test $ π (x) ge log log x $ for $ x> e ^ {e ^ 3} $. There is a way to try out that also goes for $ 2 le x e e ^ {e ^ 3} $ otherwise (in the worst case) some valuable source to verify this numerically?