So?

Got any more harder ones, based user

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Try proving this one.

Is this the highschooler edition thread?

Looks like it.

>interesting
Do you get excited when doing retarded algebra manipulation? Jesus

wolframalpha.com/input/?i=(1-4i)z=10+11i

wolframalpha.com/input/?i=(1-4i)z=10+11i

It's exactly:

Z=(10+11i)/(1-4i)

Maybe I'm understanding this one wrong because I can produce counterexamples

If you're doing homework for free might as well do mine.

is this a problem in rudin?

not that I'm aware of

You need to review series from the ground up.

I pretty much solved that for you in the other thread.

You're understanding it wrong. It's asking if the SEQUENCE converges, not if the SERIES of the sequence converges. Also, why would he need to review series? I think he just misread or something.

Done!

You too, although I'm not sure if its rigorous. Tips appreciated.

My dick gets hard when I FOIL desu

>Thinly veiled homework thread

I guess I'll post some old problem sets from a class I took 2 years ago. Have fun.

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Could user break down Einstein summation Convention for someone whos not able to use or understand it?

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To be fair,Veeky Forums is not only for math,but science too.To belong in Veeky Forums and not /b/ you only should know 10th grade maths like pic related

z = (10+11i)/(1-4i) = (10+11i)*(1+4i)/17 = (-34+51i)/17 = -2 + 3i.

z=(10+11i)/(1-4i)

multiply by complex conjugate of denominator to simplify.

Get 2+3i=z.

This is physics tho, user

Then it should be easy :^)

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Am I just retarded or is this quite simple?

This is just the average. Put the first sequence tail within epsilon and we can put the second within delta as the earlier terms over n approach 0. They converge to the same number.

Reorder your statements. Ideally sentences don't start with "since" as this means you went result->proof. Only do that when the proof is longer than one sentence.

No, no, no, absolutely not. There is an n in the sum, so you can't distribute it into the sum.

you dont know how complex numbers work do you

t. engineers

But that won't give you the step-by-step solution unless you pay for it

a_n = 1 for all n

Then a^bar_n can't converge, right

Intuitively, it makes sense because you are just taking an average of the more and more terms. It'll just converge to the same thing as the original sequence, but it gets kinda gross to do with rigorous limit proofs.

Oh, I'm stupid, I mixed up n and k

Does anyone else find the complex number notation really ugly?
Expressing a number as a + bi usually looks like shit.
I much prefer
[math]re^{\di\theta}[/math]

Does anyone else find the standard complex number notation really ugly?
Expressing a number as a + bi usually looks like shit.
The polar notation does look better, even if it's a bit more difficult to use.
[math]re^{i\theta} > a + bi[/math]

and checked

Call the limit "a" Given e=epsilon, there is a positive integer M such that |a-a_n|M. There is an integer N>M such that 1/n sum_{m=1}^M |a_m| < e/2 when n>N. Etc.

You dun goofed. 3n-1 became 3n+2 in the next step.

>theoretical comsci MSc student
>can't solve any of this shit
Am I fucked?

complex numbers can't have a greater-than relationship, user.

Uhh yeah they can, in a 2d vector space where radius is the metric

If you're fine with shit like -5 > 2, I guess.

I shifted the bottom of the infinite sum from n=1 to n=0 to apply the infinite geometric series formula.

>tfw were in accelerated math class and skipped 2 years
>still didnt leartn this before grade 11

norway is so fucking shit, its education system treats everyone like they're literally retarded. I wish i was home schooled so fucking much as i'd be working wit ayy lmaos by now.

These are actually pretty spicy

The mobile app gives a step by step solution. It's only on desktop that you need to pay.

you can use 1 though

But I didn't

just read the wikipedia article for that. if you dont understand just remeber to hide the summation sign and sum over the same index. If someone can point out how to use latex on Veeky Forums again i could write up a better example

z = (10 + 11i)/(1-4i) obviously

fucking brainlets, i stg

wolframalpha.com/input/?i=(1-4i)z = 10+11i

If you passed calc 2 you should be able to solve this

That means no wolfram alpha