Recent content by Samme013

The exercise is listed under a chapter which is before determinants are introduced( i have no clue what they are yet) so most likely it is not expected to come up wit that solution.I will definitely come back and take a look at this when i get to the next chapter though, thanks for helping !

I am not sure if i can use that fact and i can't really ask but it is kind of trivial if i can use it.Is it even possible to show it without using that?

OK so i can prove that the given inverse is actually the inverse but i can not prove that I+A is non singular without using the given inverse so how do i go about doing that?(I have done part a) Thanks in advance.

Wow that was kinda simple , i def need more practice manipulating factorials , thanks for the help man.

Yes that is as far as i go but how is that equal to the statement for n=k+1

Ok i did a and n b i proved it for n=0 assumed for n=k and used the fact that Γ(x+1)=xΓ(χ) when calculating Γ(χ+1+ 1/2) but could not find a way το prove that it is equal with what one gets just by plugging n=k+1 in the statement that we want to prove

Need help on exercise 2 from the linked image , left first in so you guys could see the Γ(χ) function any help is appreciated , thanks in advance!

Ok so i got step (a) and found that $\int_{0}^{\infty} \,d (x^0)*e^(-ax)dx=1/a$ But i do not get how i should go about starting the next steps using the info from the first step(have not done a similar problem before so i need to get a grasp on the method)

for u=1/x , du=-1/x^2 dx so substituting into the given integral and changing the integration limits sice x=a -> u=1/a and when x-->0 u--->-infiniy since x approaches 0 from the left sice in the given integral a is the other integration limit and is negative so: $f(a)=-\int_{1/a}^{-\infty} \...

Just tried something else and got a different answer: by substituting u= 1/x and if we get : $ f(a) = \int_{-\infty}^{1/a} \,d e^t ft = e^(1/a) = [e^(1/a)]/a^2]$ ... a = 1 or a = -1 so a = -1. Did i do something wrong or is it not possible to take partial derivatives like we did before since a...

Derp yeah thought a was positive , that was the other root i got , thanks !

Awesome , i took the partial derivative with respect to a on both sides and got two answers , 1 posiive , a = $\sqrt{2} - 1$ is that right?

Thanks !, i tried that but i get the following: $\ \lim_{{c}\to{\infty}} \int_{c}^{1/a} \,e^t dt = f(a)=\lim_{{c}\to{\infty}} e^(1/a) - e^c $ Which is just -infinity and does not help me find a

Ok so i am stuck on this problem , i tried substituting 1/x for u did not help , tried turning it into a limit where the top integration limit would be c and c-->0 did not get anywhere.Also tried substituting f(x) for its anti derivative [-e1/x] with no luck.So yeah at least some tips to get me...

Need help with this exercise been stuck on it for a while i think i get the gist of what i am supposed to do but can't seem to get it to work i am definitely missing something. I set h(x)= x-2/3 - g(x) and tried using the mean value theorem on [a,b] and then tried finding the minimum value of...