Recent content by FallArk

Duh... Of course! How did I forget! Thank you!

I don't think it is possible, because IVT requires that for some d (in this case 3) between g(a) and g(b), there is a point c where g(c) = d. but the function g(x) gives the value of 1/2 and 2 on [1,2] Did I miss something? Maybe it could work and I forgot something important

I don't think Rolle's theorem would work though, then g(1) would need to be equal to g(3), but they are not... - - - Updated - - - if x is not 0, and 0 at 0. Thus f'(x) exists for all x but since the limit of f'(x) as x goes to 0 is not 0, f'(x) is not continuous at x= 0. I know it does not...

One more question, we did not learn that property in class. Is it possible to prove that f' is a continuous function? so I can use the intermediate value theorem to say that f'(c) = 1/2 exists

Prove that if f is a differentiable function on R such that f(1) = 1, f(2) = 3, f(3) = 3. There is a c $$\in$$ (1 , 3) such that f'(c) = 0.5 I think the mean value theorem should be used, but I can't figure out how to prove such value exists

I found this on stack exchange https://math.stackexchange.com/questions/60340/fibonacci-modular-results

I wanted to prove that the Fibonacci sequence is a divisibility sequence, but I don't even know how to prove it. all I know is that $$gcd\left({F}_{m},{F}_{n}\right)={F}_{gcd\left(m,n\right)}$$ and I should somehow use the Euclidean algorithm?

$$L\left(f,{P}_{n}\right) = \sum_{i=1}^{n}\left(1-\frac{i}{n}\right)\cdot\frac{i}{n}$$ then take the limit

Like this? $$\lim_{{n}\to{\infty}}L\left(f,{P}_{n}\right) = \lim_{{n}\to{\infty}} \left(1-\frac{i}{n}\right)\cdot\frac{1}{n}$$

Then the lower bound of the upper sum would be $$1+\frac{i-1}{n}$$ After that do I simply evaluate the sums?

What would be the standard partition? $$P= \left\{[{x}_{0},{x}_{1}],...,[{x}_{i-1},{x}_{i}],...,[{x}_{n-1},{x}_{n}]\right\}$$ ? I think 1-x where x is irrational would be the lower sum, and it should be bounded above by 1, since 0 is not irrational. And 1+x would be the upper sum bounded by 1 as...

Prove that the function $$f(x) = 1+x, 0 \le x \le 1$$, x rational $$f(x) = 1-x, 0 \le x \le 1$$, x irrational (they are one function, I just don't know how to use the LATEX code properly) is not integrable on $$[0,1]$$ I don't know where to start, I tried to evalute the lower and upper Riemann...

Ooooo, that is clever! and just so happen that $$U(f,{P}_{n})$$ is almost the same since $${M}_{i}$$ is $$\frac{i}{n}$$! which further proves that this function is integrable

I ran into some issues when trying to calculate the lower Riemann sum of $$f\left(x\right)={x}^{3}$$, $$x\in[0,1]$$ I am asked to use the standard partition $${P}_{n}$$ of $$[0,1]$$ with n equal subintervals and evaluate $$L(f,{P}_{n})$$ and $$U(f,{P}_{n})$$ What I did: $$L(f,{P}_{n}) =...

I was so concentrated on getting rid of the h, I did not even see that I can just evalute it. Thanks! - - - Updated - - - Thanks! I get it now