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 realise 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 Îµ>0 there exists a N such that for all n>N the following inequality...
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...
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...