Recent content by laurabon

sorry if I am re opening this , I have the same problem , can you explain why $\mu$ is the product $\alpha\times \sigma$, where $\alpha$ is given by $d\alpha= x^{n-1} d\lambda$.? Really thanks

In the maximum modulus therem we have two sets $E1={z∈E:|f(z)|<M}E2={z∈E:|f(z)|=M}}$ I know that the set E1 is open because pre image of an open set. But should't be also E2 closed because pre image of only a point ?

I came up with this . Is correct? ## \begin{aligned} & \left|f(u) e^{i \omega(t-u)}\right|=|f(u)| \\ & \iint|f(u)|=\int_{-\infty}^{\infty} c = \infty \end{aligned}## how should I use the fact that sin and cosine are not integrable ?I mean to verify that is in L1 I always have to use the fact...

it is Fuorier inverse theorem with ##f## in ##L^1(\mathbb{R})##

Hi everyone in the following expression ##f(t)=\frac{1}{2 \pi} \int\left(\int f(u) e^{-i \omega u} d u\right) e^{i \omega t} d \omega ## the book says I can't swap integrals bacause the function ##f(u) e^{i \omega(t-u)}## is not ## L^1(\mathbb{R} \times \mathbb{R})## why ? complex...

Yes you are right , it was my error . I was considering the denominator of taylor expansionfor log(1+x) with factorial but... it doesn't hae factorials... about what you proposed in yyour post , doing multiplication shouldn't a_0 be 0? because on the right I have 1+0x+0x^2=a_0x

yes. Is it wrong ? the coefficient for log are different because I have used the Talor series but the calculation is that

I agree with you and mabe is the safest way . About my calculation : I have used also walphram alpha and it says that 1/(x-x^2/2+x^3/6) (I used only these terms for ln(1+t) = 1/t+1/2+t/12−t^2/24 instead 1/log(1+t)=1/t+1/2-t/12+t^2/24 why this happens?

I have tried to use the natural Taylor expansion of ln(t+1) and working with long division but I get the result 1/t+1/2+t/12−t^2/24+⋯ instead of 1/t+1/2-t/12+t^2/24+... that is the right result I have tried to do this several time but still don't works. Is there a miscalculation in my long...

Hi , thanks for your replies . Can I use the fact that 1/(1-x)=1+x+x^2+... and so 1/(1+x^2)=1-x^2+x^4+... and obtain f'(0)=-1 f^3(0)=-2! and so on ?

I have to write taylor expansion of f(x)=arctan(x) around at x=+∞. My first idea was to set z=1/x and in this case z→0 Thus I can expand f(z)= arctan(1/z) near 0 so I obtain 1/z-1/3(z^3) Then I try to reverse the substitution but this is incorrect .I discovered after that...

my question is how can I approch the problem ? And what is explicitly the set f(Eα)? {f(x) ∈ [a, b] such that what ??}

the first method is this : I think I can create a surjective function f:[0,1]^n→S^n in this way : [0,1]^n is omeomorphic to D^n and D^n/S^1 is omeomorphic to S^n so finding a surjective map f is equal to finding a surjective map f':D^n →D^n/S^n and that is quotient map. Now if I take now a...

I think that I found the correct solution to this question . The possible minimal polynomials are $$(x+1)^a(x+2)^b$$ with any $$0≤a≤2, 0≤b≤2, 1≤a+b≤3$$ now what about characteristic polynomial ?I only need this to find the jordan form am I right?

Thanks , next time i'll use them