Recent content by mariama1

Thanks for your notes , but the f(x) seems to be not uniformly continuous , but I am not sure

Yessss, I understod now thanks a lot for your good advices . I solved it

Sorry,but i do not know :(

and how can i show that ? can you help me ?

Homework Statement Show that this function f_{n}(x)= x^{5}+nx-1 has exactly one real zero point and it is in the interval \left(\frac{1}{n+1},\frac{1}{n}\right) Homework Equations By calling the zero point a_{n} decide if the series \sum \left(-1\right)^{n-1} a_{n} converges...

The taylor serries for sin (x) = \sum\frac{-1^{n}}{(2n+1)!}x^{2n+1} and by substituting (pi x^2)\2 we get \sum\frac{-1^{n}}{(2n+1)!}\left(\frac{\pi\times x^{2}}{2}\right)^{2n+1} Woow It is a power series now s(x) =\int \sum\frac{-1^{n}}{(2n+1)!}\left(\frac{\pi\times...

Homework Statement Find the power series representation for s(x) and s`(x) integral sin (pi t^2)\2 and which of them is valid ? Homework Equations The Attempt at a Solution I tried to solve this question , but i am not sure s`(x) = sin (pi t^2)\2...

Thanks Yes , i think you proof is right because x arctan x seems to be uniformly cont. because it is bounded between 90, -90 but about cos ln x , if we take the limit of f`(X) = - sin 1\x the limit when x goes to 0 does not exist . i think it is enoygh to prove that this function is...

1- the first one if we find f`(x) = 1\x and if we find the lim when x goes to 0 , then the limit does not exist So , the function is not uniformly cont. on this interval right ? but how can i solve the next one ?

determine if these functions are uniformly continuous :: 1- \ln x on the interval (0,1) 2- \cos \ln x on the interval (0,1) 3- x arctan x on the interval (-infinty,infinty) 4- x^{2}\arctan x on the interval (infinty,0 5- \frac{x}{x-1}-\frac{1}{\ln x} on the interval (0,1)...

Can you complete the answer please ?? I am interesting to know how we can prove this .