Wonderful formula: I've now finished an algorithm that calculates cosines and sines in equal steps with this formula. You give cos(x) resp. sin(x). It calculates all cos(2x) .., cos(3x) ... cos(nx) resp. sin(nx) for a given n without trigonometric functions. Could be useful for sound production or application of Fourier series.
For a moment I forgot you were reuploading your bprp content here and thought "hey wait isn't this that geo by geometric sum?" Where its the sum of various geometric sums which differ geometrically.. I guess evaluating 1cos1 + 3cos2 + 6cos3... (n(n-1)cos(n)/2) would be relatively trivial
Um, did you make a sign error in the numerator when doing the quotient rule in the beginning? I think you might have turned a minus into a plus, but maybe I'm wrong.
Are you sure that's right? Using a value of N=1, the result isn't cos(1). I think you erred in the quotient rule, where you added the two terms in the numerator. ie you did vu'+uv'.
I'm sorry to interrupt you, Dr. Peyam, but in a video that you did some months ago, you solved a complex integral using a Bromwich Contour, and I wanted to ask you if you could tell me how is that video called, because I'm learning how to solve improper integrals using that contour and I can't get it yet, so I was hoping if I could see that video again, I could finally get it. Greetings and please keep with the awesome videos!!
Any function from set A to set B can be continuous, IF you make A and B into topological spaces (and they're suitably compatible). The details are a bit complicated, but a topology on a set A means you've chosen subsets of A that are called "open"; the collection of all open subsets of A is called the topology of A. An example from the real numbers R that might help is the "standard topology", where the open sets are (basically) open intervals (a,b). That's actually where the word "open" comes from. Topologies do have some conditions - e.g. the empty set must be an open set - but they aren't super important here. Once you have a topology for the set - and topologies can be very wild - you've got a topological space. A function (or map) between two topological spaces f: A -> B is continuous if the pre-image of every open set in B is open in A. Don't worry about pre-images if you aren't familiar - this just means that the topologies on A and B are somewhat compatible, and that f doesn't drastically alter any open sets from either space. A continuous function f: R -> R is one that doesn't rip apart open intervals too much, i.e. doesn't take close points to faraway points. But what topology can Z have? There are probably many possible ones, but there's a natural one that comes from Z being a subset of R, the subset topology. Basically you use the topology on R to build the one on Z. Specifically, if U is an open set in R - you can think of it as an interval (a,b) - then U intersect Z is open in the subset topology on Z. Now that Z has a topology, you can talk about continuous functions f: Z-> Z. By the way, this topology on Z turns out to be the discrete topology, meaning that EVERY subset of Z is open, making it... not especially interesting. In this case, every function f: Z -> Z is continuous. But like I said, there are other possible topologies on Z. The standard topology on R can be defined via the metric |x-y|; "metric" means a function that measures distance between two elements. The absolute value of their difference is the most natural way, and you can define open sets in R via this metric, leading to the standard topology. But if you use a different metric, you can obtain other interesting topologies. Number theorists like to use the "p-adic" metric, which defines two integers to be close together if their difference is divisible by a high power of a prime number p. There's a p-adic metric for every prime number. E.g. 150 and 22 are quite close in the 2-adic world, while 3 and 2 are quite far apart, because 150-22=128 is divisible by 2^7, while 3-2=1 is only divisible by 2^0. The p-adic metrics can be used to obtain a topology on Z that's quite interesting. Importantly, the choice of topology on the sets in question will change the meaning for continuous functions between those sets. My point, besides telling you about some cool math, is that "continuous" means nothing unless you know the topologies you're dealing with.
+himanshu mallick Actually, you don't need to introduce topological spaces to see that EVERY function f:Z→Z is by definition continuous. You just need to apply the exact definition of a function f to be continuous at a point x=a. The definition mostly found in books is: A function f:D→R (where D - the domain of f - is a subset of R) is continuous for x=a, if for every ε>0 (however small) there exists a δ>0 such that if |x-a|
So, I plugged n = 100 into the equation and it turns out quite different. There's a mistake in there somewhere. just with a simple sum from 1 to 100 the decimal approximation is -3.378612653. Plugging n = 100 into the formula you have gives -3.143686230
okay, but how does that affect the final formula? I've tried changing the signs in the final formula and it still does not work. Maybe I should go through it and try to fix it.
@@gnikola2013 That restriction applies to infinite series, not finite sums. In this case the only restriction is that x not be a multiple of 2π (the common ratio of the GS can't be 1).
Allow me a suggestion. Before recording the video write the result of the problem to be solved. After recording the video, make sure the recording result is correct. This will avoid too many write errors. thank you.
At 3:12 it’s a minus sign, not a plus sign!
Wonderful formula: I've now finished an algorithm that calculates cosines and sines in equal steps with this formula. You give cos(x) resp. sin(x). It calculates all cos(2x) .., cos(3x) ... cos(nx) resp. sin(nx) for a given n without trigonometric functions. Could be useful for sound production or application of Fourier series.
~ 1:30 e^(nix) = 1
What's the matter?
Is nix 0 in german?
PackSciences why do i remember you from somewhere...
nix is like the short form of "nichts", which translates to "nothing" so it's basically 0.
Excellent presentation sir .Thanks.DrRahul Rohtak Haryana India
Like the new enlarged white board! Another really interesting video.
Great job!!! Keep doing more fun math videos, cos(i)
For a moment I forgot you were reuploading your bprp content here and thought "hey wait isn't this that geo by geometric sum?" Where its the sum of various geometric sums which differ geometrically.. I guess evaluating 1cos1 + 3cos2 + 6cos3... (n(n-1)cos(n)/2) would be relatively trivial
Very beautiful result you showed...I love it..very clever and smart senpai
Sensational. Thanks for posting.
very ingenious. I first thought of using the double angle formula, but I did not realize i could use de Moivre's theorem
Complex numbers are magical.
Agreed
Um, did you make a sign error in the numerator when doing the quotient rule in the beginning? I think you might have turned a minus into a plus, but maybe I'm wrong.
Kenny Lucas im sure he took minus in the top ;)
Are you sure that's right? Using a value of N=1, the result isn't cos(1). I think you erred in the quotient rule, where you added the two terms in the numerator. ie you did vu'+uv'.
I'm sorry to interrupt you, Dr. Peyam, but in a video that you did some months ago, you solved a complex integral using a Bromwich Contour, and I wanted to ask you if you could tell me how is that video called, because I'm learning how to solve improper integrals using that contour and I can't get it yet, so I was hoping if I could see that video again, I could finally get it. Greetings and please keep with the awesome videos!!
Look at all the videos about integral of 1/1+x^n or 1/1+x^4 or sin(x^2)
@@drpeyam thank you so much! You're the best teacher ever!!
More complex analysis !!)) Thanks.
0:56 so the last term is always 1?
Awesome! Though it's a special case of the Lagrange trigonometric identity. Maybe it would be nice to see the general case.
I read somewhere "let f be a continuous function from Z to Z..."
How can such as function be continuous?
Any function from set A to set B can be continuous, IF you make A and B into topological spaces (and they're suitably compatible).
The details are a bit complicated, but a topology on a set A means you've chosen subsets of A that are called "open"; the collection of all open subsets of A is called the topology of A. An example from the real numbers R that might help is the "standard topology", where the open sets are (basically) open intervals (a,b). That's actually where the word "open" comes from. Topologies do have some conditions - e.g. the empty set must be an open set - but they aren't super important here.
Once you have a topology for the set - and topologies can be very wild - you've got a topological space. A function (or map) between two topological spaces f: A -> B is continuous if the pre-image of every open set in B is open in A. Don't worry about pre-images if you aren't familiar - this just means that the topologies on A and B are somewhat compatible, and that f doesn't drastically alter any open sets from either space. A continuous function f: R -> R is one that doesn't rip apart open intervals too much, i.e. doesn't take close points to faraway points.
But what topology can Z have? There are probably many possible ones, but there's a natural one that comes from Z being a subset of R, the subset topology. Basically you use the topology on R to build the one on Z. Specifically, if U is an open set in R - you can think of it as an interval (a,b) - then U intersect Z is open in the subset topology on Z.
Now that Z has a topology, you can talk about continuous functions f: Z-> Z. By the way, this topology on Z turns out to be the discrete topology, meaning that EVERY subset of Z is open, making it... not especially interesting. In this case, every function f: Z -> Z is continuous.
But like I said, there are other possible topologies on Z. The standard topology on R can be defined via the metric |x-y|; "metric" means a function that measures distance between two elements. The absolute value of their difference is the most natural way, and you can define open sets in R via this metric, leading to the standard topology. But if you use a different metric, you can obtain other interesting topologies. Number theorists like to use the "p-adic" metric, which defines two integers to be close together if their difference is divisible by a high power of a prime number p. There's a p-adic metric for every prime number. E.g. 150 and 22 are quite close in the 2-adic world, while 3 and 2 are quite far apart, because 150-22=128 is divisible by 2^7, while 3-2=1 is only divisible by 2^0.
The p-adic metrics can be used to obtain a topology on Z that's quite interesting. Importantly, the choice of topology on the sets in question will change the meaning for continuous functions between those sets.
My point, besides telling you about some cool math, is that "continuous" means nothing unless you know the topologies you're dealing with.
@@Bignic2008
Thank you very much for shedding light on this wonderful piece of math, which I was unaware of!
+himanshu mallick
Actually, you don't need to introduce topological spaces to see that EVERY function f:Z→Z is by definition continuous. You just need to apply the exact definition of a function f to be continuous at a point x=a.
The definition mostly found in books is:
A function f:D→R (where D - the domain of f - is a subset of R) is continuous for x=a, if for every ε>0 (however small) there exists a δ>0 such that if |x-a|
this is amazing...,I guess there is a small quotient rule error,the formula isn't outputting 1 for cos(1)
Brilliant
I believe, on 3:12 it should be minus, not plus. But anyway, awsome video like all the others on your chanel :) Привет из России
Yeah, that's realy true, your formula doesn't work according to my calculator :(
When applying the quotient rule it should appear a minus sign!!
So, I plugged n = 100 into the equation and it turns out quite different. There's a mistake in there somewhere.
just with a simple sum from 1 to 100 the decimal approximation is -3.378612653.
Plugging n = 100 into the formula you have gives -3.143686230
See the pinned comment
okay, but how does that affect the final formula? I've tried changing the signs in the final formula and it still does not work. Maybe I should go through it and try to fix it.
1+e^ix+e^2ix+...+e^nix = (e^i(n+1)x -1)/(e^ix -1)
But shouldn't this only hold when |e^ix|
|e^ix| = 1 whenever x is real
@@drpeyam yeah but shouldn't it be less than one?
@@gnikola2013 That restriction applies to infinite series, not finite sums. In this case the only restriction is that x not be a multiple of 2π (the common ratio of the GS can't be 1).
@@MichaelRothwell1 oh of course I mixed things up somehow, thanks
please a lecture on complex analysis....contour intgrl...
Already done ✅
Dr. Peyam's Show ...thnks...sir....i like ur videos..great job ..
This is a recursive formula: If you have cos(1)..cos(n) you can find cos(n+1) //Square(Sin(1/2)) is a constant
Maybe you could further simplify 4*sin^2(1/2) into 2(1 - cos(1))
at 2:49, when u derivate e^(inx) it would be ne^(i(n-1)x) in place off ne^(inx)?
Don't confuse the differentiation of exponentials like e^x with powers like x^n. The power reduces by 1 for x^n, not for e^x.
Can you do a video solving cos(1/1) + cos(1/2) + ... + cos(1/n) ?
sir you used plus sign instead of minus sign in using quotient rule.
alright thanks for watching, with comment.
You Gotta admit that there are
No dislikes.
*ILLUMINATI CONFIRMED*
6:38 I did not get the logic here... Could you explain? Thanks
Oops, nevermind. I just realized I should have kept watching :)
3:12 it should be minus, not plus , right?
Allow me a suggestion. Before recording the video write the result of the problem to be solved. After recording the video, make sure the recording result is correct. This will avoid too many write errors. thank you.
I’m already doing that
2:32 How is it guaranteed that if you differentiate the members you get another true equation?
the two members are different representations of the same function of x, so the derivatives of both sides must be equal
Notice that N is finite, so you dont even have the problem of differentiatin a inffinite series
Got it, thanks for your answer
There was an error while writing the derivative of e power inx.Also there was sign error while writing the quotient rule.
Killing two birds with one TRIG (As in like trick but its also short for trigonometry)... ill go now 😭
Dr. Peyam, everytime I watch one of your videos I try to imagine your IQ... What's your score, exactly? XD
Last time I checked it was 💯
At 10:04 couldn’t you just take real terms
that's a bigger whiteboard
I dont know about this 😪
Setting N = 1 we find that the result is WRONG!
You really made a mistake when you find (f/g)’
See pinned comment