How to show function is converging towards a point or not.

Say F is function of n and x. Say F=(n*x)/1+(n*x)^2 Is given function then for x[0,k] where k belongs to (0,1). Show that when n tends to infinity will give converging function. Using graphs. Basically just take limit as n tends to infinity where x belongs accordingly. Please I need code as I am getting few error.

[madhan ravi]

Upload your code ,x[0,k] what does this mean? Is it a function?

[T]

No generally when you find limit on variable x say we find for all values for x but here x belongs to 0<=x<=k.( closed interval [0,k] Where k belong to open interval 0<k<1. Please help me out as I am getting error like how we will define x and k together. Then find limit for n tends to infinity with conditions on x . Using graph we need to observe.

[Walter Roberson]

I suspect that you are missing () in F. No one bothers to write a division by 1,but people do accidentally omit ( for division by (1+something)

[T]

F= (n*x)/(1+(n*x)^2)

[T]

Now find limit when n tends to infinity where x has condition that it will take value between closed interval [k,1] and where k belongs to open interval (0,1) That all I need graph for that particular limit to have clear picture. Please help me with code as I am getting few error.

[madhan ravi]

Where is your code that you tried?

[T]

 sysm n,x
X=0:k;
K=(0:1);
y = fun of n and x.
Limit((y),n,inf)
Plot(x,n,y)

Something like this

Answer 1

[madhan ravi]

syms n x k
assume(0<k<1) %openinterval
assumeAlso(0<=x<=k) %closedinterval 
F= (n.*x)./(1+(n.*x).^2) 
limit(F,n,inf) 
F=@(x,n) (n.*x)./(1+(n.*x).^2)
x=linspace(0,1,1000);
n=1:1000;
plot(F(x,n))

[T]

I did with assume syntax also but getting error And plot(F,n,x) will be fine or not

[madhan ravi]

Yea the difficult part is the graph now , I am figuring it out

[T]

Yes please that will be great bcs once we will have graphs then everything will be clear through graph whether it is converging or not

[madhan ravi]

See the edited code now

[T]

Sure did you check on matlab I mean is it working

[madhan ravi]

Yes it’s working and the graph moves towards zero

[madhan ravi]

see attached screenshot of the graph

[T]

Thanks a lot. it is possible to have these graphs with 3d view

[madhan ravi]

syms n x k
assume(0<k<1) %openinterval
assumeAlso(0<=x<=k) %closedinterval 
F= (n.*x)./(1+(n.*x).^2) 
limit(F,n,inf) 
F=@(x,n) (n.*x)./(1+(n.*x).^2)
x=linspace(0,1,1000);
n=1:1000;
F= (n.*x)./(1+(n.*x).^2)
plot3(x,n,F)
xlabel('x')
ylabel('n')
zlabel('F')
grid on

The above will display it in 3d , see attached screenshot

[T]

Thanks a lot, Dear thanks for helping me.

[T]

If you don't mind can you please tell me how to have proper command on Matlab.BCS it is very useful to have proper understanding.

[madhan ravi]

What do you mean by proper command?

[T]

proper command I mean how to do master in Matlab

[madhan ravi]

see lots of videos and there is suggested books in mathwork website https://in.mathworks.com/academia/books/mastering-matlab-hanselman.html

[T]

n=1:1000; & x=linspace(0,1,1000); used for what

[madhan ravi]

n starts from 1 to 1000 and x from 0 to 1 divided into 1000 parts

[T]

That I know we are taking less value BCS to have a graph properly contained ri8.

[madhan ravi]

yes according to the assumptions (conditions)

[T]

also, we used . while writing function BCS it is a way to express when we do multiplication

[madhan ravi]

yes it's element wise operation , each element of x is multiplied with corresponding element of n