Neat and simple. I suppose one simple thing to add would be the intuition behind the Cauchy product, namely that you could, say, put the terms in a multiplication grid, and the Cauchy product simply adds all terms in the grid by going along the diagonals.
For any solution to f’(x) = f(x)^2, f(x + c) is also a solution, and the set of all solutions is a one parameter family, so the only functions with this property are 1/(c - x)
@@silver6054 Oh oops! Though if you “let c be infinity” and accept my lack of rigor you do get that, and the one parameter family is now topologically a circle, which is pretty cool.
I subscribed when I first saw your video on Lorentz transformations, I appreciate the level of detail you put into your explanations and your clarity stands out.
all the way I was thinking ... that's not going to work ... that's not going to work ... that's not going to work ... that's not going to work .... and then .... ooohhh it DOES work 😂
Wow! I just came across the piano recital on You Tube from four years ago. I was mispronouncing your first name. I can't understand why I seem to have a mental block on infinite series. Maybe it'll come to me one day. I have the time now to learn this stuff. Years ago, I found out why so many of us can't get math, so I could blame my professors, but I think it's a third the student, a third the text, and a third the professor. I'll always remember that in ten minutes how you explained LaPlace transforms that I could understand, 😀without the "piecewise continuous" confusion and the lack of explaining how the integral works I had back in the fall of 1976, which left me confused for nearly a half century.😵💫
Is it possible to construct a function whose half-order derivative is equal to the sum of its second and third prior result? f(x) {'...' }(1/2) = f(x-2)+f(x-3) ?
It is better to have a look at geometric aspect of this. Suppose a square with length 1+x+x^2+x^3+.... and try to evaluate the area of the square by its induced partitions.
0:30 No, I don't agree. E.g. take x=5 than 1/(1-x)=-0.25 but the sum of all powers will diverge towards infinity. More worse: take x=0 and you will "prove" infinity=0 👎
@@MuPrimeMath If you mentioned "-1 < x < 1" at the start of the video it would have saved me scrolling down here to find this comment! Thanks though :)
If it's true, lol, and I'm guessing it is, otherwise the video wouldn't say it is, lol...write out the partial sums, etc, do the operations on the partial sums, take the limits, then establish they are the same, etc...
Say you have a finite sum Sn = 1 + q + q² + q³ + ... + q^n Then, by distributive property, q*Sn = q + q² + ... q^(n+1) If you subtract Sn from q*Sn, you get a telescoping series, which results in q^(n+1) - 1. (Apply commutative property to aggregate each equal term on both series) Simply divide each side by q and you get the result of the original finite sum. Sn = (q^(n+1) - 1)/q Now, it happens that if |q|
Perhaps you have seen how to solve 1+1/2+1/4+1/8+1/16+... That sum is the first value (1) over 1 minus the common ratio (1-1/2) or 2. That formula is 1/1-r = 1+r+r^2+r^3+... This only converges when -1
Neat and simple. I suppose one simple thing to add would be the intuition behind the Cauchy product, namely that you could, say, put the terms in a multiplication grid, and the Cauchy product simply adds all terms in the grid by going along the diagonals.
First proof nice and immediate. Second an interesting counting procedure.
For any solution to f’(x) = f(x)^2, f(x + c) is also a solution, and the set of all solutions is a one parameter family, so the only functions with this property are 1/(c - x)
Well, also the trivial functions f(x)=0 which isn't of that form
@@silver6054 Oh oops! Though if you “let c be infinity” and accept my lack of rigor you do get that, and the one parameter family is now topologically a circle, which is pretty cool.
Your explanation makes things easy to grasp.👏
I subscribed when I first saw your video on Lorentz transformations, I appreciate the level of detail you put into your explanations and your clarity stands out.
Only applicable when |x|
It also works when x = 0.72927499488311371218371090973371
@@Sir_Isaac_Newton_also applicable when x = 0.972762946715444173900173976153849912547988155025108
@@Sir_Isaac_Newton_The absolute value of a number between and not including 0 and 1 is obviously less than 1
Uh, why ?
@@briogochill6450 Its a rule for binomial expansion of negative and decimal index.
Otherwise series will always coverge to infinity.
Great explanation. World class. Thank you.
Ok, this time I subscribed to your channel. First time that I understand the Cauchy thing. ☺
2:29 Microphone distortion leads to EX-CUTE 😂
Amazing result, I feel delighted to be your early subscriber
I understood it yippiiiieeee. Great explanation
That is beautiful.
all the way I was thinking ... that's not going to work ... that's not going to work ... that's not going to work ... that's not going to work .... and then .... ooohhh it DOES work 😂
Only if abs(x) < 1
1/(1-x) isn't continuous
Wow! I just came across the piano recital on You Tube from four years ago. I was mispronouncing your first name. I can't understand why I seem to have a mental block on infinite series. Maybe it'll come to me one day. I have the time now to learn this stuff. Years ago, I found out why so many of us can't get math, so I could blame my professors, but I think it's a third the student, a third the text, and a third the professor. I'll always remember that in ten minutes how you explained LaPlace transforms that I could understand, 😀without the "piecewise continuous" confusion and the lack of explaining how the integral works I had back in the fall of 1976, which left me confused for nearly a half century.😵💫
Thanks for making this video
Hey bro it's 3:45 am here good morning 😅
thanks dude for this video
Is it possible to construct a function whose half-order derivative is equal to the sum of its second and third prior result? f(x) {'...' }(1/2) = f(x-2)+f(x-3) ?
very nice explanation! more like this please :)
It is better to have a look at geometric aspect of this. Suppose a square with length 1+x+x^2+x^3+.... and try to evaluate the area of the square by its induced partitions.
Then you have to derivate another square, but how!?
Very good.
Isn't d/dx((1-x)^-1) =-(1-x)^-2
No, because the negative signs from the power rule and from the derivative of 1-x cancel out.
Cool result
d/dx(1/1-x)=1/(1-x)^2...(1/1-x)^2=1/(1-x)°2
Seulement pour série infinie
Sinon
1+2x+3x² ≠ (1+x+x²)²
0:30 No, I don't agree. E.g. take x=5 than 1/(1-x)=-0.25 but the sum of all powers will diverge towards infinity.
More worse: take x=0 and you will "prove" infinity=0 👎
The infinite sum equation holds for |x|
@@MuPrimeMath If you mentioned "-1 < x < 1" at the start of the video it would have saved me scrolling down here to find this comment! Thanks though :)
x
Equation of friction?🤔
If it's true, lol, and I'm guessing it is, otherwise the video wouldn't say it is, lol...write out the partial sums, etc, do the operations on the partial sums, take the limits, then establish they are the same, etc...
you forgot to mention that the absolute value of x must less than 1..
Exactly I was also thinking the same thing
@@AbhinavKumar-nh8dli was also thinking of saying I was thinking that this is exactly what I was thinking of saying
Not necessarily, if you treat the series as formal power series, you don’t have to worry about convergence
nice video i like the huh? cat
Why would the coefficient be n+1
∫ (Σ ♾️ /n=0 x^n)•dx =(Σ•1/2• ♾️)^2
(Σ1/2• ♾️ )^2 = [( ♾️ /2)^2+c]
[ ♾️^2 /4+c]=[♾️ +c]
dx is a square matrix [[0, 1], [0, 0]]
-2/x = S(xⁿ)
Use S(G.P.) formula for x < 1 and solve. Since if x > 1 then LHS -> -1
RHS -> ±infinity
Hence x < 1
Why is the sum equal to 1/1-x? If i sub in x it clearly is not equal
It is equal only if -1 < x < 1
Say you have a finite sum Sn = 1 + q + q² + q³ + ... + q^n
Then, by distributive property, q*Sn = q + q² + ... q^(n+1)
If you subtract Sn from q*Sn, you get a telescoping series, which results in q^(n+1) - 1. (Apply commutative property to aggregate each equal term on both series)
Simply divide each side by q and you get the result of the original finite sum. Sn = (q^(n+1) - 1)/q
Now, it happens that if |q|
Perhaps you have seen how to solve 1+1/2+1/4+1/8+1/16+...
That sum is the first value (1) over 1 minus the common ratio (1-1/2) or 2. That formula is 1/1-r = 1+r+r^2+r^3+...
This only converges when -1
🇩f=f²......
you can try to do the derivative of the geometric series formula with respect to x AND n.
Stoopid cat... 😁
Clean shaven fits you better bro. Embrace your youth when you still have it.