example, constant line is convex and concave %%visits: 3
strictly concave if f(θx1 + (1 − θ)x2) < θf(x1) + (1 − θ)f(x2)
convex if f(θx1 + (1 − θ)x2) ≥ θf(x1) + (1 − θ)f(x2)
Convexit := convex downwar := Let f ∈ C(a, b)
f is called convex on
(a, b), if for any
x1, x2 ∈ (a, b)
and for any θ ∈ (0, 1) we have
f(θx1 + (1 − θ)x2) ≤ θf(x1) + (1 − θ)f(x2)
f is called concave on (a, b) if (−f) is convex on △
properties of convex functions - they are cts - Lipschitz_continous in a compact interval - f″ monotonicity
If f″ > 0 then you are convex
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For any 2 points in the interval, The graph between two points is always less than a straight line between the two points.
concave is the same but for the function −f
point_of_inflectio := If f is convex on (x0 − δ, x0)
and concave on (x0, x0 + δ)
then x0 is called
the point of inflection
also
point_of_inflectio := If f is concave on (x0 − δ, x0)
and convex on (x0, x0 + δ)
then x0 is called
the point of inflection
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Let f be twice differentiable on (a, b). - If f″(x) > 0 on (a, b),then f is convex on (a, b) - If f″(x) < 0 on (a, b) then f is concave on (a, b)
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If f″(x) > 0 on (a, b),then f is convex on (a, b) Proof:
Let x1, x2 ∈ (a, b) with x1 ≠ x2. Define g ∈ C[0, 1]
g(θ) = f(θx1 + (1 − θ)x2) − θf(x1) − (1 − θ)f(x2), θ ∈ [0, 1]
we want to show g ≤ 0 as this will prove convexity. Direct calculation: g(0) = g(1) = 0 and $$ g''(\theta) = (x_1-x_2)^2f''(\thetax_1 + (1-\theta)x_2) $$
suppose for contraition sake, g > 0 somwhere on [0, 1]
Maximum theorem: g attains its (positive) maximum at θ0 ∈ (0, 1) Fermats_Theorem ⟹ g′(θ0) = 0 By our assumption, g″(θ0) > 0 so it is a local minimum. So θ0 is a local minimum and local maximum of g. Only possible if g = constant but then g″ = 0 and g′ = 0 which is a contradiction.
Suppose every week z take home a salary, Sn, so. $$ \sum _{k=1}^{10} S_n = 100 $$
Suppose get a bonus
so Sn → Sn2
$\frac{1}{10}\sum _{k=1}^{10} S_n \le \frac{1}{10}\sum _{k=1}^{10} S_n^2$
Suppose I present an investment portfolio. Each year my revenue is R(n) at week n Sps. $\sum _{k=1}^{S_2} R(n) = 10^{10}$ usually If $R(n) \mapsto R(n)^{\frac{3}{4}}$ What is the best case scenario for $\sum _{k=1}^{S_2}R(n)^{\frac{3}{4}}$
Wts $\sum _{k=1}^{S_2} X_n$
We know $\sum _{k=1}^{S_2}= 10^{10}$
Example
F is convex
Youngs inequality = for A, B ≥ 0, p, qst $\frac{1}{p} + \frac{1}{q} = 1$
$AB \le \frac{A^{p}}{p} + \frac{B^{q}}{q}$
Chose a,b so ea = Ap, eb = Bq
$AB = f^{\frac{a}{p}} + \frac{e^{b}}{q}$
sps I need to sum $| _{n=1}^{} a_n b_n | $ which is oscillatory
sps (a1…, an) and (b1, …, bn)
and
we know that $\left| \sum _{n=1}^{\infty} a_n b_n \right| \le \left( \sum _{k=1}^{K} \left| a_k \right|^2 \right)\left( \sum _{k=1}^{K} \left| b_k \right|^2 \right)$ which is not as
Sps i want to compute the area of infinite boxes
$\sum _{k=1}^{\infty} \left| l \times h \right|$
If $\sum _{k=1}^{\infty} |l|^{4} = 1$ and $\sum _{k=1}^{\infty} |h| = 1$
Then $\sum _{k=1}^{\infty} \left| l \times h \right| \le 1$
y = x2 ## resources tags :math: