Fermat's unique method of integration.
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- Опубліковано 28 сер 2023
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History of math, both of things the people did, as well as the math they developed, is always an interesting dive.
For the second integral, you wrote it wrong in the beginning. Should've been 1/x^n and you just put x^n
With regards to the n=1 case for negative exponents, I would love to see your take on historical methods for finding the “quadrature of the hyperbola”!
In the vein of integration before calculus, pascal's method of integrating the sine and cosine is fascinating and requires only geometry
Where might one find a discussion or an account of such a method?
@@19divide53 Dirk Jan Struik's book "A Source book in Mathematics, 1200-1800" has a whole load of fascinating pieces of math history. This particular example is on page 238
13:56 Oh so that time, the margin was large enough for that proof?
This wasn't about number theory, so it wasn't written in the margin of Diophantus' book. ;)
So damn cool!! I first encountered this as an undergraduate in a history of mathematics course (back in the 70's). Just as remarkable (a nod to Fermat) is his method of finding tangents to polynomials as well as Descarte's method for finding tangents - double roots. Worthy of a video? All pre-Newton. Bloody marvelous. Long live infinite geometric series.
This looks like the reverse operation to the q(uantum) derivative?
I was thinking the same - but the quantum derivative arrived only in the early 20th century. May be some delving into the history might show something similar to the quantum derivative much earlier, an obscure bit of Euler perhaps...?
Fun to watch - to get an inkling of what Fermat thought - and entirely sensible
This was so interesting and intuitive.
Very interesting. Nicely illustrates that the limit has to be the same regardless of the sequence of simple functions.
Thank you! 🙏🙏🙏
Did you also know that any infinite series that you can prove converges by the ratio test intuitively tells you something very remarkable. Such a series behaves more and more and more like an infinite geometric series the further and further you go out in the series. The ratio |a(n+1) / a(n)| as n --> inf. approaches the magic value of |r| < 1 for an infinite geometric series. Long live convergent infinite geometric series.
In one sense yes, and in another sense no. In the sense you mean, only the "tail" of the series behaves more and more like a geometric series, while the values of the series are much better approximated by the values up to a finite order approximation. If we consider our series to take complex inputs, then the radius of convergence tells us how far we are away from a singularity, where either the (tail of the) series behaves like 1/z^n or like exp(-z). The series for exp(z) has an infinite radius of convergence, while the standard geometric series converges to (1-z)^-1 which blows up at z=1. This comparison shows that even different tails can exhibit different properties, since no finite order approximation really captures the behavior of the series near z=1.
Very nice! The masters of the past had their own methods...
This is genius. Even if it you'd have to prove it 'converges' to the true value of the integral (as given by the limit of Riemann sums), it's still a very ingenious idea.
Love your content
Good stuff! Can you make a video about Donald Knuth’s approach, where he uses Bachmann/Landau notation to define derivatives? Thanks!
Partitioning problems :) I would blindly give Fermat my disk for partitioning and be happy with the sector ranges :)
Isn't there a negative sign missing in the second integral result?
No, because a is on the lower bound
@@Alex_DeamAha, I didn't see that.
Amazing!
Wow, didn't know about this one :)
You could also just take the limit of (1-q)/(1-q^(n+1)) using L'Hopital's Rule to get 1/(n+1) without bothering with the factorisation. Not sure that's any simpler though, both quite simple.
L'Hopital would be useful to Fermat but unfortunately it didn't exist yet. L'Hopital's rule was only discovered in 1694, well after Fermat's death in 1665
Wiki says a mathematician named Cavalieri used geometry to prove all positive integer cases.
Really cool that Fermat used this method, I suppose it was based on Archmedes 'method of exhaustion'? He probably also computed the lower bound of inner rectangles to ensure they got the same value.
So the general formula would be that ∫ f(t) dt = lim_{q->1} a (1 - q) (f(a) + f(aq) q + f(a q^2) q^2 + .... )
I wonder if there's any integrals that this makes easier than the "standard" Riemann sum. It seems like the extra q factors make it not work out nicely unless f power function.
For example, if we want to integrate e^t we get a (1 - q)(e^a + q e^(aq) + q^2 e^(aq^2) + ... ) and I'm not sure that you can do much with that
If we define e^x as the limit of (1+x/n)^n, then we might consider f(n,x)=(1+x/n)^n.
Let g(n,x)=x^n, then f(n,x)=g(n, (x+n)/n). By general properties of integrals (as areas), its integral on (0,t) is 1/n^n times the integral of g(x+n). But that's just the difference of the results from Fermat's method for x^n by taking x=t+n and x=n. Therefore, integral of f(n,x) on (0,t) is ((t+n)^(n+1)-n^(n+1))/(n^n(n+1)) = ((t+n)*(1+t/n)^n-n)/(n+1).
Since t is a finite number, (t+n)/(n+1) tend to 1 as n approaches infinity. n/n+1 tends to 1 as well, and (1+t/n)^n tends to e^t by definition. So the integral tends to e^t-1. In other words, the integral of e^x on (0,t) is e^t-1.
5:43 this actually corresponds to the LIMIT of the sum of the geometrical series that otherwise would be 1-r^n instead of 1 in the numerator. This is not really mentionned and it is not entirely rigourus if I am understanding well... I guess we would retrieve the same expression in the end but I think it is worth mentionning.
You get the same expression back, which should be expected since 1-q^(n+1) = (1-q)(1+q+q^2+ ... + q^n) identically for all q by polynomial division/multiplication depending on the direction you're working. You see the same possibility around 12:49 with the finite geometric series in the denominator. However, you need to cancel the (1-q) in either case so that the sum can be evaluated as q→1.
There is another approach of this integral which makes use of the notion that the integral of x^n from a to b equals the area under the curve. If it is area, it must satify the basic properties of area (linearity, additivity, scale invariance, translation invariance and normalization of area, ie the integral of 1 from 0 to 1 must be 1. When these conditions are met, we will find the integral from 0 to a of x^n equals 1/(n+1) a^(n+1). One can proof this without any limiting proces, just finite calculations, for all rational numbers.
Do you have a link to an explanation of that method?
Yes I have: ua-cam.com/video/Js2mwsHc4p4/v-deo.html
If we define I(n,t) to be the integral of x^n dx on (0,t) for n>=0, and similarly the integral of x^n on (t, infinity) for n
'and the same integral ..'?? I take it that you try to extend the method Wildberger explained for n being a real number. As real numbers themselves are the result of an infinite process, I don't expect a finite process for such an integral. A less ambittous extension would be if you try to extend it for all integers n. However, with the integral of 1/x from a to b you run already into a problem. The result is not a rational number and therefore can only be approximated with with finite processes. I will think about this a bit more.
There is already 1/4 of a million of subscribers :) - it makes 1/28000 people on earth are interested for example of Fermat way of integration :) (instead of the lame Gaussian cubatures and quadratures that everybody knew about ;)
A question: In the second part we are working on (1/x^n), so why on the right part of the blackboard the integral is written for x^n ? Thanks.
Exactly, trouble at 8:56, just add a minus sign to the n exponent 😊
Can a similar approach be used on functions other than x^n? In this case a sequence of partition points X_n = (a, q a, q^2 a, ... ) is used to generate the rectangles. For other functions f(x) you would have to use a different sequence of partition points. So can this method be generalized?
Well you could possibly get the Taylor series of said function and integrate it
Is this the inspiration for q-analogs?
And also can you help whit the other integration fórmulas ?
PLEAS
Is this the first video where peach-colored chalk has made an appearance?
can you do it in same way but for sin(x)?
7:52 q>1
Pls who knows how he factored out 1-q in the denominator at 6:28
Hello,
Mi name is Josafat and please I need you yo make a video about the integral of e^x
I try to bild these resultados but I don't know how yo start
That was mentally stimulating, even out of the box. 🙂 But I don't see any advantage of geometric partitioning of the rectangles.
Fermat's integration method "is like maybe" Quantum / Greek
Want to know more about Fermat
Does the method work for fractional exponents? If so, how?
proof fractional exponents via implicit differentiation
Yes, nothing he did assumed n was an integer.
Fermat was the real inventor of The Differential Calculus!
Huh? This was about integrals, not about differentials.
@@bjornfeuerbacher5514: I'm aware of that but just wanted to point that out!
Too bad Fermat was a jurist and not a full time mathematician!
He was a lawyer. I bet it paid much better
mate its not about pay in life@@TheEternalVortex42
Slippin fermat