Recent content by Felafel

Hello everyone! I'm a civil engineering (bachelor) student, and I was fascinated by the "hydraulics" course. unfortunately, my study plan doesn't include other courses on the matter for at least one year. Thus, I am looking for some easy books to begin with, to study it a bit on my own...

when you say do you mean I should study the convergence of the associated series afterwards, and see if this happens? yes, i didn't realize it could be negative (case x<1), so i should have changed the inequalitiy's signs and examine the two different cases: x<1 and x>1 separately do...

i'm sorry it's taking so long, but i really don't seem to get this topic, which shouldn't even be that difficult, after all. i'll give another try, more carefully. i want to see for which ns this inequality holds: ##|ln(1+x^{1/n}+n^{-1/x})-ln2|<\epsilon ##. it is the same as...

ok, i'll also do a short recap to see if everything's in order. - I calculated the pointwise convergence and got that the function goes to ln2. - to find the uniform convergence i have to prove that for every x and for every ε>0 there exists a N such that for all n>N the following inequality...

ok! Now it's clearer. So, convergence is not uniform in |R, but how do I find out if there are subsets of |R where the convergence is uniform?

doesn't it hold for any x?because the logarithmic function goes to infinity more slowly than n. Shoul I assume from that that it is uniformly convergent in all |R+?

do you mea that i have to find n s.t. ##1/n<log_x(1+\epsilon)##? And for ## n>\frac{1}{log_x(1+\epsilon) ## it converges uniformly? From which passage did i get that expression?

any further hint? i really can't make it :(, i had the following results but they seem absurd to me. i tried to put: ##sup|f_n(x)-f(x)|=| sup(f_n(x)) - inf f(x)|= |sup (ln(1+x^(1/n)+n^(-1/x))| - ln2| = sup | ln(\frac{1+x^(1/n)+n^(-1/x))}{2}|## now if x=1 i get ##lim |ln(1/2+1/2+1/n)|=ln1=0## so...

i've tried to do as you suggested: there exists an ε > 0 such that for every natural number N there exists x ∈ S and and n ≥ N with |fn(x) − f(x)| > ε ##|ln(1+x^{1/n}+n^{-1/x}-ln(2)| > \epsilon## ##|ln\frac{1+x^{1/n}+n^{-1/x}}{2}| > \epsilon## if x=1 ##|ln\frac{1+1+n^{-1}}{2}| > \epsilon...

sorry again, they are sequences. yes, i was referring to weierstrass' m-test. can't I use it for sequences too?

you are probably right in saying convergence is not uniform, but ii really keep not seeing why my method doesn't work. i'll try to write things a bit differently. by the deginition of uniform convergence i have to prove there exists a n big enough such that: ##|f_n(x)-f(x)|< \epsilon##...

and, as far as the second series is concerned, I've just computed what follows: ##|\frac{x}{n} e^{-n(n+x)^2}| ## = ##|\frac{x}{n} \frac{1}{e^{-n(n+x)^2}}| \leq |\frac{x}{n} \frac{1}{1+n(n+x)^2}| ## given ##e^x \geq x+1## ##\leq |\frac{x}{n} \frac{1}{n(n+x)^2}| ## dividing num and denom by x i...

also, if there aren't values of x that go on well with the definition of uniform convergence, can I say that the function converges uniformely in any compact subset of |R for the aforemetioned reasons (beginning of the thread)?

isn't it actually false even for x<1? if i get x=1/2, for instance, ##(1/2)^{1/n}+n^{-2}## goes to 1+0 if n is big. so all the values x can have involve a contradiction with the definition of uniform convergence, as the ##sup|f_n(x)-f(x)|## isn't less than epsilon

sorry, I've made a bit of a mess, I'm trying to correct myself: 1) f_1n convergese pointwise to ln(2) as previously said, but I'm not studying the uniform convergence differently: according to the definition, for n big enough i get: ##|f_n(x)-f(x)| < \epsilon## ##\forall \epsilon >0##, ##\forall...