By checking with Wolfram Alpha the answer for the fourth power seems to be (pi*sinh(pi))/(-cos(sqrt(2)*pi)+cosh(sqrt(2)*pi)). I need a video explaining how you find that.
Before i finish the video, i think the infinite product will be 0, because you have an infinite amount of numbers below 1, making each product slightly smaller over and over, thus approaching 0.
@@tfg601 well, it turns out i was wrong, but that is what a limit is. Infinity isn't a number you get to, its a process. ever step contains finite numbers, we will never get to infinity. By the end of the process we will have .6666 because everything cancels out. don't think of it as literally infinity, think of it as what happens as the numbers get closer to infinity, they never will be infinite.
A fun tip for next time is that convergence of infinite products can very easily be reduced to convergence of infinite series. Step 1. If you have negatives you're either going to 0 or doomed to diverge so you might as well look at the absolute value. Step 2. Take the logarithm of the product. That product becomes a sum, and the terms just become their logarithms. Step 3. Your terms can almost always be reduced to 1+1/(something that goes to 0), do that and then use the taylor series of ln(1+x), and the sum's convergence is now very easy to prove. Conclusion : If it diverges to -infinity, then the product probably goes to 0. If it diverges to +infinity, then the product does as well. If it goes to some value, then the product might go to the exponential, it might be close, or it might not be at all, but we know it converges. The proof to all of that is a bit more complicated but yeah, if you have the infinite product of 1 + something(n), it has the same convergence as the sum of something(n), except in the case of -infinity where it goes to 0 instead. Very useful to evaluate things like this product on the go, because the sum of -2/(n^3+1) very obviously converges so that product converges to something we know is not 0.
Finally, a fast converging series for computing 2/3 !
Lmao 🤣. That was spoiler haha.
@@jixpuzzle Haha, you're right, sorry for the spoiler 😂
The type of math that makes you smile.
Was about to say that :)
this channel is great :0 :0
thank you so much for making these !!! :)
By checking with Wolfram Alpha the answer for the fourth power seems to be (pi*sinh(pi))/(-cos(sqrt(2)*pi)+cosh(sqrt(2)*pi)). I need a video explaining how you find that.
I like this channel
Before i finish the video, i think the infinite product will be 0, because you have an infinite amount of numbers below 1, making each product slightly smaller over and over, thus approaching 0.
I was wrong =)
NO THAT DOESN'T MAKE SENSE, BECAUSE THEN IT WILL BE INFINITY/INFINITY
@@tfg601 well, it turns out i was wrong, but that is what a limit is. Infinity isn't a number you get to, its a process. ever step contains finite numbers, we will never get to infinity. By the end of the process we will have .6666 because everything cancels out. don't think of it as literally infinity, think of it as what happens as the numbers get closer to infinity, they never will be infinite.
@@localidiot4078 yeah I know 👍
A fun tip for next time is that convergence of infinite products can very easily be reduced to convergence of infinite series.
Step 1. If you have negatives you're either going to 0 or doomed to diverge so you might as well look at the absolute value.
Step 2. Take the logarithm of the product. That product becomes a sum, and the terms just become their logarithms.
Step 3. Your terms can almost always be reduced to 1+1/(something that goes to 0), do that and then use the taylor series of ln(1+x), and the sum's convergence is now very easy to prove.
Conclusion : If it diverges to -infinity, then the product probably goes to 0. If it diverges to +infinity, then the product does as well. If it goes to some value, then the product might go to the exponential, it might be close, or it might not be at all, but we know it converges.
The proof to all of that is a bit more complicated but yeah, if you have the infinite product of 1 + something(n), it has the same convergence as the sum of something(n), except in the case of -infinity where it goes to 0 instead. Very useful to evaluate things like this product on the go, because the sum of -2/(n^3+1) very obviously converges so that product converges to something we know is not 0.
"one step"
Lol
Step 1. Find it.
I love this kind of math.
Very easy question, bog standard method of telescoping over an infinite product
you live in a bog
Criminal how this cannel only has 12400 subs
keyword: telescoping product
In the start, you took the value of n from 2, why not 1?
well if n was 1 we would have had 0/2 as our first term and anything multiplied by 0 would be 0, thus a very boring answer
So underrated
Do you study computer science or math ?
Both, both is good
Nice video
nice video!
2/3
Easy, 1^3 -1=0, therefore, the answer is 0
Except there's no such term called (1³-1)/(1³+1)
Answer = 0
Only correct if it starts from (1³-1)/(1³+1)
However we're starting from (2³-1)/(2³+1)
Etc is pronounced et-cetera, not exedra (but nice video for the rest :)
It’s just his accent lol
shuddup grammer notsee!!!
@@tompeled6193 Nah, it's genuinely good advice