How do you solve the system #4xy + 4y^2 - 3cos(y) = -2#, #5x^2 - 2xy^2 - 5sin(x) = -5#?

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Answer 1 · 2016-09-10T14:22:06

See below.

Using an iterative process.

Making #f(x,y) = ((4xy+4y^2-3cos(y)+2),(x^2-2xy^2-5sin(x)+5))#

Calling #p = (x,y)# and #p_0 = (x_0,y_0)#

The linear expansion around #x_0,y_0# is

#f(p) approx f(p_0)+grad f_(p_0)(p-p_0)#

Supposing that #abs(f(p)) approx 0# we have the iterative procedure

#p = p_0 - (grad f_(p_0))^(-1)f(p_0)# or

#p_(k+1) = p_k - (grad f_(p_k))^(-1)f(p_k)#

Here

#grad f = ((4 y, 4 x + 8 y + 3 sin(y)),(2 x - 2 y^2 - 5 cos(x), -4 x y))#

Starting with #p_0 = (2.32703,-2.11049)# we obtain the convergent sequence

#((1.70737,-1.42624), (1.15774,-1.01564), (1.00499,-0.939189), (1.00591,-0.942905), (1.00591,-0.942895))#

within #6# significant digits.