Recent content by SclayP

Isn't a way to prove this without a counterexample. The injectivity it's all right if i doi it that way, but to prove that the function it it surjective i have that: \forallyB,\existsxA/y=f(x)

So, i have to analyze the next function and see if it is injective and surjective f: NXN → N f(x,y)=x+y i know that a function to be injective f(a)=f(b) → a=b but here i have that f(a,c)=f(b,d) → (a,c)=(b,d) next thing i do: a+c =b+d now i don't know what to do...PLEASE HELP ME

1.Find n, if the term 11 coefficient it is 6 time the term 10 coefficient in 2.(6x^7 + 5x^(-4))^n 3

[b]The problem statement says to find out if the next series converge, and if it does to calculate the sum with an error ε< 10^{-3} The serie is this one \sum^{\infty}_{n=1} (-1)^nne^{-n} First of all the serie converges because of Leibniz Criterion but the i did the series of |an| I did it...

Oh! hahaha i can't believe it was that, and i can't believe i didnt notice while doing it...well thank you for you help... PD: I know it's late but...sorry for my english

Yes, that is what I GET but calculator says that the resault it is 2...and what we get by doing it "by hand" it 4.19 so, there is a problem in the proceedure.

well yes you are right be we ended up with the same resault because ln(x^2 -1) equals ln[(x-1)(x+1)] equals ln(x-1) + ln(x+1)... but knowing that we came to the same and assuming it's all correct why is the length not the same that the one that give my calculator...

Yes, i'v already done that and then i integrated by part the integral \int\frac{v}{(v-1)(v+1)} and i come up with this \frac{v^2}{2} |^{\sqrt{e^2 +1}}_{\sqrt{2}} + \frac{1}{2} \int^{\sqrt{e^2 +1}}_{\sqrt{2}}\frac{1}{v-1} + \frac{1}{v+1}\, dv So when i integrate 1/(v-1) and 1/(v+1) it's left...

I don't need one i just say t=e^{u}+1 ; dt=e^{u}du and so du=\frac{dt}{e^{u}}=\frac{dt}{t-1}

In the first place i wouldn't know how to do it, and in the other hand i don't think it's necesary i mean, i think the way i did it it's correct but I'm stuck in that step. I used the next formula L=\int \sqrt{(\frac{dy}{dx})^{2} + 1}

Sorry, you're right the funcion is f(x)=e^x between 0 and 1

So I'm stuck resolving this integral... \int^{1}_{0} \sqrt{e^{2x} + 1} \, dx u=2x ; du=2dx \frac{1}{2}\int^{2}_{0} \sqrt{e^u + 1} \, du t=e^u+1 ; dt=e^u du \frac{1}{2}\int^{e^2 + 1}_{2} \frac{\sqrt{t}}{t-1} \, dt v=\sqrt{t} ; dv=\frac{1}{2\sqrt{t}} \int \frac{v^3}{v^2 -1} \, dv And here i...

So the integral does not converge after all...

I actually wrote the wrong limits of integration but the math i did it right. i just wrote it wrongly,but yes it would be \int^{\frac{\pi}{2}}_{0}csc(x^2) \, du = -\frac{1}{2}ln(csc(x^2) + cot(x^2)) + C |^{\frac{\pi}{2}}_{0} Then \lim_{b\rightarrow 0} {-\frac{1}{2}ln|csc(x^2) + cot(x^2)| + C...

YES, your absolutly right..Im getting it from Wolfram the thing is that what I'm writing is the equivalent for resticted u values. I'm sorry. So... \int^{\frac{\pi}{2}}_{0}csc(u) \, du = -\frac{1}{2}ln(csc(u) + cot(u)) + C |^{\frac{\pi}{2}}_{0} Then, \lim_{b\rightarrow 0}...