Recent content by renolovexoxo

I hope this is in the right place, it feels like calculus, but it's the last part of my analysis problem. Construct an example where g: R2->R lim x->a g(x) exists but lim ||x||->||a|| g(x) does not exist I'm having a very hard time coming up with something to put this together. I think...

Error using inlineeval (line 15) Error in inline expression ==> 2.71828^x-ln(2)*2^(-1*x)-2*sin(x) Undefined function 'ln' for input arguments of type 'double'. Error in inline/subsref (line 24) INLINE_OUT_ = inlineeval(INLINE_INPUTS_, INLINE_OBJ_.inputExpr, INLINE_OBJ_.expr); Error in...

That didn't work. I am still getting the same error.

Okay, that worked great! Thank you. Last question, just because I have a hard time in Matlab. When I'm looking at the solution, do I want to display c or f(c) to get the root? It would be c, correct?

I've been using this for a Newton Approximation in Matlab function x = Newton(f, fp, x, nmax, e) % f is an inline function which we apply Newton's method on % fp is an inline function that is the derivative of function f % x is the initial guess of the root % nmax is the total number...

Here is the code I have, but I keep getting the error message: Undefined function 'f' for input arguments of type 'double'. I don't know what I have that is causing this. Does anybody see what's wrong with my code? MaxIt = 1000; epsilon = 10^-5; a=1; b=2; c = (b+a)/2; NumIt = 0...

Thank you for the help, but you don't need to make me feel bad for not understanding something you see is obvious. I do appreciate you pointing it out to me, I was having hard time seeing if it was the composite version.

Homework Statement I need to answer the attached question, but we haven't done anything similar in class and the book isn't proving to be helpful for me. I'm not sure if I have to use matlab, we were given this code but I can't even understand what it is calculating. % f(x), the function...

No, It is when I move to evaluate it for x=1.3. I was given x values for 1.1, 1.2, 1.3, and 1.4. When I evaluate 1.3, what is the range of possible y values that I can consider? Does it stay between (1.1, 1.3) or do I move to include the points I used in my derivative estimation?

I'm sorry, I meant e^x, I copied it from somewhere else and didn't catch the format change. I'm sorry about that! I want to know (h^2)/3 * e^x, so the error was defined by h^2/3 * e^1.3

So we are estimated derivatives using a three point formula in class and the giving bounds for the error. I was given a table of x1=1.1, x2=1.2, x3=1.3, x4=1.4. I have to find an error bound to estimate the error for x=1.1 and 1.3 for f(x)=ex In class we did x=1.1 as follows...

Thank you. That makes a lot more sense now.

I'm sorry, but I still don't really understand what it is saying. I've read it several times and really am having a hard time. We didn't go over any examples of anything like this in class.

So in the case, what would I consider k and j? Would i let n=k to finish this proof?

I'm having trouble with how I am supposed to prove it without knowing the f(n) that is being used. Is there some part of the theorem that I'm missing that would help me with that aspect of this proof?