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...
Homework Statement
study the pointwise and the uniform convergence of
##f_{n1}(x)=ln(1+x^{1/n}+n^{-1/x}## with ##x>0## , ##n \in |N^+}## and ##f_{n2}(x)=\frac{x}{n}e^{-n(x+n)^2}## with ##x \in \mathbb{R} ## , ##n \in }|N^+}##
The Attempt at a Solution
1) first series: ##f_{1n}##
studying...
thank you. so, apparently i can't be much explicit
i'll go for another try:
##\frac{\partial ^2 f}{\partial s^2}=\frac{\partial f}{\partial x}\left( \frac{\partial f}{\partial x}\frac{\partial x}{\partial s} \right)+\frac{\partial f}{\partial x} \left( \frac{\partial f}{\partial x}\frac{\partial...
Homework Statement
let u=f(x,y) , x=x(s,t), y=y(s,t) and u,x,y##\in C^2##
find:
##\frac{\partial^2u}{\partial s^2}, \frac{\partial^2u}{\partial t^2}, \frac{\partial^2u}{\partial t \partial s}## as a function of the partial derivatives of f.
i'm not sure i'm using the chain rules...
ok, i've tried to complete it:
##x+y=2u ; x-y=2t## ##\Rightarrow##
##f(u,t)=-sinu*\phi(t)## if ##t \neq 0##
##f(u,t)=-sin(u)## if ##t=0##
##\lim_{t \to 0} f(u,t)=-sin(u)## which equals f(u,t) if t=0.
Thus the function is continuous for t=0.
Also, it is overall differentiable, because it is a...
like:
x+y=2u
x-y=2t
and using the formula lurflurf suggested:
##f(t,u)=-sinu \frac{sint}{t}## if ##(t,u) \neq (0,1)## ##\rightarrow ## ##f(t,u)=-sinu*\phi(t)##
?
i'm afraid i don't really understand the logic and the text of the problem, i'm just going for random attempts.
could you please...
let me see if i've understood. i should rewrite:
##f(t+y, x-t)=\frac{cos(t+y)-cos(x-t)}{t}## if ##t \neq 0##
##f(t+y, x-t)=-sin(t+y)## if ##t=0## ##\Rightarrow## ##f(t+y, x-t)=-siny##
and then study the differentiability for t=0, which is the only "critical point", as sin and cos are...
Homework Statement
Let ##\phi## be defined as follows:
##\phi(t)=\frac{sint}{t}## if ##t \neq 0##
##\phi(t)=1## if ##t = 0##
prove it's derivable on ##\mathbb{R}##
now let f be:
##f(x,y)=\frac{cosx-cosy}{x-y}## if ##x \neq y##
##f(x,y)=-sinx ## in any other case
express f as a...
Is it correct to say that, being semidefinite positive for ##x \geq1/2## the function is convex and has therefore infinite minima and being semidefinite negative for ##x \leq -1/2## it is concave and so all the points to the left of (-1/2, 0, 0) are maxima?
Homework Statement
find the minima and maxima of the following function:
##f:\mathbb{R}^3 \to \mathbb{R} : f(x,y,z)=x(z^2+y^2)-yx##
The Attempt at a Solution
after computing the partials, i see ∇f=0 for every point in the x-axis: (a, 0, 0)
The Hessian is:
( 0 0 0 )
( 0 2a -1...
yes, i know that, but ny problem is that the differential doesn't exist if i consider xy as a single variable, but it does if i consider x and y separately, please look at the two different applications of the definition of differentiability i wrote