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saul85 [17]
3 years ago
6

Show that the sum of two concave functions is concave. Is the product of two concave functions also concave?

Mathematics
1 answer:
s2008m [1.1K]3 years ago
3 0

Answer:

1.

Let f and g concave functions. Then for a in (0,1), f( (1-\lambda)x + \lambda y ) \geq (1-\lambda)f(x) + \lambda f(y) and g( (1-\lambda)x + \lambda y ) \geq (1-\lambda)g(x) + \lambda g(y)

Now, (f+g)( (1-\lambda)x + \lambda y ) = f( (1-\lambda)x + \lambda y ) + g( (1-\lambda)x + \lambda y )\geq (1-\lambda)f(x) + \lambda f(y) + (1-\lambda)g(x) + \lambda g(y) = (1-\lambda)(f(x)+g(x)) + \lambda (f(x)+g(x))= (1-\lambda)(f+g)(x) + \lambda(f+g)(x)

This shows that the function f+g is concave.

2. Let f and g concave functions.

(fg)( (1-a)x + \lambda y )= f( (1-\lambda)x + \lambda y )g( (1-\lambda)x + \lambda y )  \geq [(1-a)f(x) + \lambda f(y)] [(1-\lambda)g(x) + \lambda g(y)] = (1-\lambda)^2 f(x)g(x) + \lambda (1-\lambda) f(x)g(y)+ \lambda (1-\lambda) f(y)g(x)+ \lambda ^2 f(y)g(y) \neq (1-\lambda)f(x)g(x) + \lambda f(y)g(y)

This shows that the product of two concave functions isn't concave.

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A sequence is defined recursively by the following rules: f(1)=3f(n+1)=2⋅f(n)−1 Which of the following statements is true about
Radda [10]

Answer:

f(2)=5

f(5)=33

Step-by-step explanation:

The given formula, that recursively defines the sequence is

f(1) = 3 \\ f(n + 1) = 2f(n) - 1

When n=1, we obtain;

f(1+ 1) = 2f(1) - 1 \\ f(2) = 2 \times 3 - 1 \\ f(2) = 6 - 1 \\ f(2) = 5

When n=2, we get:

f(2+ 1) = 2f(2) - 1 \\ f(3) = 2 \times 5 - 1 \\ f(3) = 10 - 1 \\ f(3) = 9

When n=3,

f(3 + 1) = 2f(3) - 1 \\ f(4) = 2f(3) - 1 \\ f(4) = 2 \times 9 - 1 \\ f(4) = 18 - 1 \\ f(4) = 17

When n=4

f(4 + 1) = 2 f(4) - 1 \\ f(5) = 2 \times 17 - 1 \\ f(5) = 34 - 1 \\ f(5) = 33

When n=5,

f(6) = 65

4 0
3 years ago
2+cotA=1 homework help
Alex Ar [27]
2+\cot A=1\implies \cot A=-1\implies \tan A=-1

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The divide and simplify of the equation √36 x⁴ / √9 x⁶ will result to  \frac{2}{x^{2} }

Let's begin by simplifying the equation;

Simplify the equation by finding the square root of both 36 and 9

The square root of 36= 6

The square root of 9= 3

= \frac{6x^{4} }{3x^{6} }

The above equation can also be written as;

=  6x^{4}  * 3x^{-6}

The next step is to solve the powers, that is,

=   \frac{6x^{4} }{3x^{6} }

Cancel out 6 by 3 and get  2

The result is,  

=  \frac{2}{x^{6-4} }

The final submission result is,

=  \frac{2}{x^{2} }

To find more on divide and simplify, go to: brainly.com/question/16356152

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