Fallenleaves
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@CountBlecka) D
b) Converges to 3/4
c) By p value test, it converges to -1/3
@CountBlecka) D
b) Converges to 3/4
c) By p value test, it converges to -1/3
a) it actually converges to 0.@CountBleck
Ogrea) it actually converges to 0.
b) it actually converges to 1/4
c) it converges to pi/4. not -1/3.
Why can't we approximate the integral of part c to be 1/x^4a) it actually converges to 0.
b) it actually converges to 1/4
c) it converges to pi/4. not -1/3.
why are you using an approximation?Why can't we approximate the integral of part c to be 1/x^4
Infinity is in the domain of xwhy are you using an approximation?
i don't understand. what does this mean?Infinity is in the domain of x
Ah I fucked up, ignore what I asked. How do you solve part c? Thanksi don't understand. what does this mean?
Ah I fucked up, ignore what I asked. How do you solve part c? Thanks
Thanks a lot, I've never seen this used before
Postmax "You're Welcome"Thanks a lot, I've never seen this used before
It's quite long, you'd still have to integrate cosine squared xPostmax "You're Welcome"
it would be a pain to overleaf anymore, but it's a simple method. you could either use IBP or just use the double angle formulaIt's quite long, you'd still have to integrate cosine squared x
Using the double angle formula seems more intuitive to meit would be a pain to overleaf anymore, but it's a simple method. you could either use IBP or just use the double angle formula
1/6 because 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6.Limit (1² + 2² + 3² + ... + n²)/n³
n---->infinity
Guys what do you think of this guy's style?
I like his style a lot, what are the types of teaching styles that you likeI only watched the first few minutes. Not my cup of tea. Too whimsical for my liking.
1/3, I think you forgot the 2n^21/6 because 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6.
You mean 2n^3. Because its the cubic term that gives 2.1/3, I think you forgot the 2n^2
2n² or 2n³, the result is the same, I just said 2n² because you can simplify it (n(n + 1)(2n + 1))/6n³You mean 2n^3. Because its the cubic term that gives 2.
If you simplify, all the other powers of n will disappear after the limit. Only 2 remains because its the coefficient of cubic term (2n^3). Because it'll cancel out with the n^3 at bottom leaving a 22n² or 2n³, the result is the same, I just said 2n² because you can simplify it (n(n + 1)(2n + 1))/6n³
Yeah I forgot the 2. Thanks for pointing it out.1/3, I think you forgot the 2n^2
While I don't think it much matters, PDEs (partial differential equations) are usually considered a more advanced topic than parametric functions in my experience. The beauty is self-study, however, is that you get to study what you want in whatever order you want. I don't think you really need one to study the other, so just go with whatever order floats your boat I would say.@Ahnfeltia Is it normal to continue to partial differentiation after learning first order ODE's?
The DE YouTube Playlist continues to partial differentiation after teaching ODE's but the book continues to parametric functions and so on before reaching the chapter about partial differentiation.
It's infinite series time babyWhile I don't think it much matters, PDEs (partial differential equations) are usually considered a more advanced topic than parametric functions in my experience. The beauty is self-study, however, is that you get to study what you want in whatever order you want. I don't think you really need one to study the other, so just go with whatever order floats your boat I would say.
Does (-1) to the power of n, summation of all terms from 1 to infinity diverges or converges?It's infinite series time baby
I haven't started yet, I'll answer this in a week or so.Does (-1) to the power of n, summation of all terms from 1 to infinity diverges or converges?
not at all what I expected, but I'm all for itIt's infinite series time baby
It's the chapter after 'Parametric Equations and Polar Coordinates'.not at all what I expected, but I'm all for it
You got one of them big calculus books then?It's the chapter after 'Parametric Equations and Polar Coordinates'.
It's the one by Stewart, it's an ebook. I've also downloaded another Calculus book but it's by Larson.You got one of them big calculus books then?
my calculus book in uni was the one by Adam & EssexIt's the one by Stewart, it's an ebook. I've also downloaded another Calculus book but it's by Larson.
I see, I wonder what differences there could be between commonly used Calculus books in universities, they must be minimal I suppose.my calculus book in uni was the one by Adam & Essex
yeah I doubt it really matters which book one usesI see, I wonder what differences there could be between commonly used Calculus books in universities, they must be minimal I suppose.
|r|=|-1|= 1>0Does (-1) to the power of n, summation of all terms from 1 to infinity diverges or converges?
It diverges, but I didn't understand your reasoning|r|=|-1|= 1>0
Therefore it is divergent.
I'm treating this as a geometric series, where r = -1 is the common ratio. Take the absolute value of r and It's larger than or equal to one (I made a mistake by writing larger than 0 in the previous post).It diverges, but I didn't understand your reasoning
Is there a better way to prove? How's this: We take the limit of (-1) to the power n as n tends to infinity which diverges (the value of the limit oscillates between 1 and -1). As the value of the limit taken is not 0, by contradiction, the series is not convergent (divergent).It diverges, but I didn't understand your reasoning
This doesn't work, harmonic series approaches zero and it's divergent.Is there a better way to prove? How's this: We take the limit of (-1) to the power n as n tends to infinity which diverges. As the value of the limit taken is not 0, by contradiction, the series is not convergent (divergent).
It isn't formal but yeah it's honest work.
ThanksThis doesn't work, harmonic series approaches zero and it's divergent.
If it's convergent, it approaches zero, but the other way around isn't true.
The better would say that it's infinite and we can't know what value it will aproach, there it diverges
here's an actual proof. i guess i used some implicitly used some lemmas but i dont careThanks
Thankshere's an actual proof. i guess i used some implicitly used some lemmas but i dont care
View attachment 692105
the easier way to see it is to look at the sequence of partial sums, which goes -1, 0, -1, 0, -1, 0, ..., which is clearly not convergent as it's alternatinghere's an actual proof. i guess i used some implicitly used some lemmas but i dont care
View attachment 692105
Is it possible for me to go into real analysis fully considering that I've barely learnt the content of infinite series and sequences in the calculus book?the easier way to see it is to look at the sequence of partial sums, which goes -1, 0, -1, 0, -1, 0, ..., which is clearly not convergent as it's alternating
@Fallenleaves
As far as I can tell, real analysis is very different from all the other math you're used to, so be prepared for that I guess. Good luckScrap this question, I'll learn both at the same time.
Thanks.As far as I can tell, real analysis is very different from all the other math you're used to, so be prepared for that I guess. Good luck