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iteration

x = 2Sin (x)

α lies between 1/2 π and 2/3 π

show that if a sequence of values given by the iterative formula

x (n+1)= 1/3(x (n ) + 4 Sin (x(n))


converges, then it converges to a root of x=2 Sin (x)

I need some guidance with this question please ( sry should have been posted in maths tsr forum)
(edited 5 years ago)
Original post by Carlos Nim
x = 2Sin (x)

α lies between 1/2 π and 2/3 π

show that if a sequence of values given by the iterative formula

x (n+1)= 1/3(x (n ) + 4 Sin (x(n))


converges, then it converges to a root of x=2 Sin (x)

I need some guidance with this question please ( sry should have been posted in maths tsr forum)


Have you ever seen a similar question before?

You're given that the formula converges, so what can you say about xn+1,xnx_{n+1}, x_{n}?
Reply 2
Avatar for Carlos Nim
Carlos Nim
OP
9
x = x/3 + 4/3 x sinx
2/3 x = 4/3 sinx = > 6x = 12sinx

its the same thing as x = 2sinx but it has been scaled to a factor of 6

is that the proof that it will converge?

about your question, yes but i cant seem to connect it to the question asked above.
each values get closer to the root ?
(edited 5 years ago)
Original post by Carlos Nim
x = x/3 + 4/3 x sinx
2/3 x = 4/3 sinx = > 6x = 12sinx

its the same thing as x = 2sinx but it has been scaled to a factor of 6

is that the proof that it will converge?

about your question, yes but i cant seem to connect it to the question asked above.
each values get closer to the root ?


The right method but I'm not sure if you would get full marks, as you dive straight in without explaining why you've got all the xns there.

The hint in my previous question was that if the equation converges, then for large values of n , xn+1 and xn will have the same value, say L. Hence the equation becomes the first line of what you posted there, and I guess it also works if it's all xn.

And yes, you can factor out the 6.
Reply 4
Avatar for Carlos Nim
Carlos Nim
OP
9
thank you so much this has been bugging me out the entire day

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