Solution of the challenge: * Decompose x^n - 1 = (x-1) * sum_{k=0}^n x^k * Apply integration by parts, carefully handling integration bounds. Notice, that induction seems only possible for the more general integral involving arbitrary lower and upper limits a and b, respectively. * Resort to Feynman's technique again in the more general context, so as to know what to prove by induction, yielding: int_a^b (x^n - 1) / ln x dx = Ei((n+1) ln b) - Ei(ln b) - Ei((n+1) ln a) + Ei(ln a), where Ei-as usual-is the exponential integral Ei(x) := PV int_{-infinity}^x e^t / t dt * Verify that the original result is obtained-by partly using L'Hospital's rule, a known expansion of Ei(x) and Tannery's theorem-for a->0 and b->1. * Apply strong induction, in the sense of using all predecessor predicates P(k), k=0,...,n, in order to arrive at P(n+1). * Do a lot of technical stuff (integrals involving Ei, manipulate (telescopic) sums, ...) in between, to make things go as you like ... ;) * Verify findings numerically. Of course, a rather intricate procedure, but very insightful, yielding a more general result as the original one, and, as always in creative maths: The path is the goal! Again, the interesting phenomenon appears, that only formulating a more general problem, seems to make the desired (induction) strategy work ... What a nice exercise, being on my holidays!
We actually don't need to check for absolute convergence here at 7:25 since the function x^y is positive on (0,1)x(0,n). Thus we can simply apply Tonelli's theorem.
At 3:45 shouldn’t the x be raised to the y+1 power? It doesn’t change anything after since x is evaluated at 1 and 0 is just x whether the exponent is 1 (like in the video) or y+1.
@bobh6728 No because you are differentiating with respect to the exponent. Also, having (y+1) as the exponent would make the resulting integral ln(y+2) + C which wouldn't be correct
@@aevmm I was referring to the step of integrating x^y dx from 0 to 1. That should give you the y+1 as the exponent and divide by it giving you (1/(y+1))x^(y+1) evaluated from x=0 to 1. 1 gives you 1/(y+1) since 1 to any power is 1. 0 gives you 0 since 0 to positive power is 0. So it didn’t matter that he had an exponent of 1, but at that point it should y+1 for the exponent. He almost missed the x completely when he paused and then squeezed it in.
A "direct approach" might be to manipulate the expression to get integral of e^x/x dx with bounds [a, a(n+1)] and you take limit as a->0. e^x can be replaced with its taylor expansion, which gives 1/x+1+x/2... But since the interval is infinitely narrow, all but 1/x disappears. So you get as the limit as a->0 log(a (n+1))-log(a)=log(n+1)
I think you can do it in a direct way (with a small caveat left to demonstrate): notation : - \int_x=a^b [...] dx integral on x from a to b - \sum_k=a^b [...] sum of k from a to b Define α ϵ C such that | α | < 1. Consider f(α) = \int_x=0^1 (x^α - 1)/ln(x) dx The change of variables y = ln(x) gives the following: f(α) = \int_y=-∞^0 (exp(αy) - 1) exp(y) /y dy Expanding as a series: (exp(αy) - 1) / y = α \sum_k=0^∞ (α y)^k/(k+1)! Making use of the common integral for an integer k \int_y=-∞^0 y^k exp(y) dy = (-1)^k \int_x=0^∞ x^k exp(-x) dx = (-1)^k k!, and inverting sum and integral (using | α | < 1) we get f(α) = α \sum_k=0^∞ [α^k (-1)^k k!]/(k+1)! = \sum_k=1^∞ (-1)^(k+1) α^k / k = ln(1+α) (analytic continuation of log on the unit open ball) Now ADMIT f is holomorphic on Re(α) >= 0. We've found an open set on which it corresponds to the analytic continuation of the logarithm, it therefore has to match it on the real positive line. We then have f(n) = ln(1+n) for any integer n. We've here calculated the required integral: - without having to go through a "trick" - still having to pass through the realm of derivatives (using an even stronger condition on the integral's regularity)
For the induction approach, you could apply the "exponential derivative" you introduced in another video to the n-th integral and check that it turns into the n+1-th integral. Now, since the derivative shifts the function by 1, you would get that the n+1-th integral is ln(n+2)
The goal of induction is to find what sequence a_n the n-th integral corresponds to. Even if applying the exponential derivative gives the (n+1)-th integral, you still don't know what the sequence a_n is, you just know that a_n goes to a_(n+1). Via induction, you need to show that *assuming the n-th integral is a_n, show that the (n+1)-th integral is a_(n+1)*, a_n in this case being ln(n+1)
@@dave-bk6vtOnly the second time. The first time, at around 4 minutes, he wrote it out and forgot the ^(y+1). I suppose most of us knew what he meant: it's a little unlikely that someone very new to calculus would be watching this.
I thought the partial derivative of x^y = (y)x^(y-1). (at 3:51) ?? (I checked Wolfram alpha its not ) - but the integral of x^y dx = (1/(y+1))x^(y+1).. So you can treat Y as a CONSTANT in integration but not in derivatives ??? confusing
There are two things happening here: One is a differentiation with respect to y (treating x as a constant). f(y) = integral[0, 1] (x^y - 1)/(ln x) dx, so taking a derivative with respect to y gives df/dy (or f'(y)) on the left and the integrand on the right becomes (x^y ln x)/ln x because x^y ⟶ x^y ln x when x is being treated as a constant (which it is, **with respect to the derivative that's happening**). The ln x parts cancel, as we get f'(y) = integral[0, 1] x^y dx. At this point, we're done with the derivative, and it's time to take the integral. But this time, we're integrating with respect to x (it's dx, after all). And in this situation, y is being treated like a constant, so we just get the normal power rule for integrals: x^(y+1) / (y+1). It all just depends which variable we're working with, and the decision was to differentiate with respect to y (since f is a function of y), then integrate with x. The fact that we're looking at the same expression x^y in both cases is just a coincidence.
I started on trying to do it by the DI method and that way is very much impractical. The derivatives of (ln(x))^-1 are kind of interesting though, I got a recurse function, C(n,k), for the nth derivative terms coefficients . d^n/dx^n (ln(x)^-1)=sum(C(n,k)x^-n ln(x)^-k) where 2
No, it is an *application* of Leibniz rule. There is a difference between the application of a rule and the rule itself. Otherwise, might as well say that Feynman's trick *literally is* the fundamental theorem of calculus, since both Leibniz's rule and Feynman's trick are consequences of the fundamental theorem.
Feynman's trick is: 1. Introduce a new variable into an indefinite integral to produce a function of that variable 2. Differentiate that function using the Leibniz rule 3. Solve the resulting differential equation to get the solution to the integral So you see it's only 1/3 the Leibniz rule...
It doesn't make sense to do this because the indefinite integral is not a function, so there is nothing to differentiate. So no, you can't directly apply Feynman's trick.
Feynman didn't invent it, he mentions that he saw it in some textbook. It just became popular because he was famous and sociable and probably spread it to others, and since it didn't have an existing name that one stuck
Of course he didn't invent it. It's just the application of the Leibniz rule. But it's popularly named after Feynman precisely because of what you just wrote
In one of his popular books, Feynman cited Fredrick S. Woods "Advanced Caculus." Feynman says his high school math teacher gave it to him. I have my dad's copy, from when he tautght at RPI in the late 50's. Sure enough, it's on page 141. Integrating under the integral is on page 145. (1934 edition.)
Solution of the challenge:
* Decompose x^n - 1 = (x-1) * sum_{k=0}^n x^k
* Apply integration by parts, carefully handling integration bounds. Notice, that induction seems only possible for the more general integral involving arbitrary lower and upper limits a and b, respectively.
* Resort to Feynman's technique again in the more general context, so as to know what to prove by induction, yielding:
int_a^b (x^n - 1) / ln x dx = Ei((n+1) ln b) - Ei(ln b) - Ei((n+1) ln a) + Ei(ln a),
where Ei-as usual-is the exponential integral Ei(x) := PV int_{-infinity}^x e^t / t dt
* Verify that the original result is obtained-by partly using L'Hospital's rule, a known expansion of Ei(x) and Tannery's theorem-for a->0 and b->1.
* Apply strong induction, in the sense of using all predecessor predicates P(k), k=0,...,n, in order to arrive at P(n+1).
* Do a lot of technical stuff (integrals involving Ei, manipulate (telescopic) sums, ...) in between, to make things go as you like ... ;)
* Verify findings numerically.
Of course, a rather intricate procedure, but very insightful, yielding a more general result as the original one, and, as always in creative maths: The path is the goal!
Again, the interesting phenomenon appears, that only formulating a more general problem, seems to make the desired (induction) strategy work ...
What a nice exercise, being on my holidays!
Of course, it should read: x^(n+1) - 1 = ..., at the very beginning ...
We actually don't need to check for absolute convergence here at 7:25 since the function x^y is positive on (0,1)x(0,n). Thus we can simply apply Tonelli's theorem.
I’m surprised you didn’t call this video “The internet’s favourite integral trick”
Frankly, if he wants to explain Feynman or integration tricks I don't give a damn if he wears clown makeup. Great video!
He’s not Fresh Toadwalker.
At 3:45 shouldn’t the x be raised to the y+1 power? It doesn’t change anything after since x is evaluated at 1 and 0 is just x whether the exponent is 1 (like in the video) or y+1.
@bobh6728 No because you are differentiating with respect to the exponent. Also, having (y+1) as the exponent would make the resulting integral ln(y+2) + C which wouldn't be correct
@@aevmm I was referring to the step of integrating x^y dx from 0 to 1.
That should give you the y+1 as the exponent and divide by it giving you (1/(y+1))x^(y+1) evaluated from x=0 to 1.
1 gives you 1/(y+1) since 1 to any power is 1.
0 gives you 0 since 0 to positive power is 0.
So it didn’t matter that he had an exponent of 1, but at that point it should y+1 for the exponent. He almost missed the x completely when he paused and then squeezed it in.
5:38 🌳
9:39 Challenge
10:03 Good Place To Stop
A "direct approach" might be to manipulate the expression to get integral of e^x/x dx with bounds [a, a(n+1)] and you take limit as a->0. e^x can be replaced with its taylor expansion, which gives 1/x+1+x/2... But since the interval is infinitely narrow, all but 1/x disappears. So you get as the limit as a->0 log(a (n+1))-log(a)=log(n+1)
I think you can do it in a direct way (with a small caveat left to demonstrate):
notation :
- \int_x=a^b [...] dx integral on x from a to b
- \sum_k=a^b [...] sum of k from a to b
Define α ϵ C such that | α | < 1.
Consider f(α) = \int_x=0^1 (x^α - 1)/ln(x) dx
The change of variables y = ln(x) gives the following:
f(α) = \int_y=-∞^0 (exp(αy) - 1) exp(y) /y dy
Expanding as a series:
(exp(αy) - 1) / y = α \sum_k=0^∞ (α y)^k/(k+1)!
Making use of the common integral for an integer k
\int_y=-∞^0 y^k exp(y) dy = (-1)^k \int_x=0^∞ x^k exp(-x) dx = (-1)^k k!,
and inverting sum and integral (using | α | < 1) we get
f(α) = α \sum_k=0^∞ [α^k (-1)^k k!]/(k+1)! = \sum_k=1^∞ (-1)^(k+1) α^k / k = ln(1+α) (analytic continuation of log on the unit open ball)
Now ADMIT f is holomorphic on Re(α) >= 0.
We've found an open set on which it corresponds to the analytic continuation of the logarithm, it therefore has to match it on the real positive line.
We then have f(n) = ln(1+n) for any integer n.
We've here calculated the required integral:
- without having to go through a "trick"
- still having to pass through the realm of derivatives (using an even stronger condition on the integral's regularity)
For the induction approach, you could apply the "exponential derivative" you introduced in another video to the n-th integral and check that it turns into the n+1-th integral. Now, since the derivative shifts the function by 1, you would get that the n+1-th integral is ln(n+2)
The goal of induction is to find what sequence a_n the n-th integral corresponds to. Even if applying the exponential derivative gives the (n+1)-th integral, you still don't know what the sequence a_n is, you just know that a_n goes to a_(n+1).
Via induction, you need to show that *assuming the n-th integral is a_n, show that the (n+1)-th integral is a_(n+1)*, a_n in this case being ln(n+1)
shouldn't the integral of x^y dx = (1/y+1)x^(y+1). ??
The antiderivative is indeed what you said, but he is evaluating it at 1 and 0. So at 0 it equals 0 and at 1, x^(y+1)=1, so you are left with 1/y+1
Yes, but he evaluated it at the same time between 1 and 0, which leads to the answer he wrote down
Answer still correct, just a bit off in the working out
shouldn't people stop polluting the internet with pointlessly obvious garbage. You know very well that step was omitted
@@dave-bk6vtOnly the second time. The first time, at around 4 minutes, he wrote it out and forgot the ^(y+1). I suppose most of us knew what he meant: it's a little unlikely that someone very new to calculus would be watching this.
Feynman’s trick always tickles my fancy.
Thank you, professor.
Isn't "a non negative integer" a longer way to say a natural number?
6:11
Is the -1 in x^y - 1 necessary?
not really. might be a typo
I thought the partial derivative of x^y = (y)x^(y-1). (at 3:51) ?? (I checked Wolfram alpha its not ) - but the integral of x^y dx = (1/(y+1))x^(y+1).. So you can treat Y as a CONSTANT in integration but not in derivatives ??? confusing
There are two things happening here:
One is a differentiation with respect to y (treating x as a constant). f(y) = integral[0, 1] (x^y - 1)/(ln x) dx, so taking a derivative with respect to y gives df/dy (or f'(y)) on the left and the integrand on the right becomes (x^y ln x)/ln x because x^y ⟶ x^y ln x when x is being treated as a constant (which it is, **with respect to the derivative that's happening**). The ln x parts cancel, as we get f'(y) = integral[0, 1] x^y dx.
At this point, we're done with the derivative, and it's time to take the integral. But this time, we're integrating with respect to x (it's dx, after all). And in this situation, y is being treated like a constant, so we just get the normal power rule for integrals: x^(y+1) / (y+1).
It all just depends which variable we're working with, and the decision was to differentiate with respect to y (since f is a function of y), then integrate with x. The fact that we're looking at the same expression x^y in both cases is just a coincidence.
... exchanging the order of integration is equivalent to the ability to take the derivative ... |- mind = blown
I started on trying to do it by the DI method and that way is very much impractical. The derivatives of (ln(x))^-1 are kind of interesting though, I got a recurse function, C(n,k), for the nth derivative terms coefficients . d^n/dx^n (ln(x)^-1)=sum(C(n,k)x^-n ln(x)^-k) where 2
By the way, this Feynman's trick is actually just Leibniz rule.
No, it is an *application* of Leibniz rule. There is a difference between the application of a rule and the rule itself. Otherwise, might as well say that Feynman's trick *literally is* the fundamental theorem of calculus, since both Leibniz's rule and Feynman's trick are consequences of the fundamental theorem.
@@DestroManiak You're right. Thanks for the clarification!
Feynman's trick is:
1. Introduce a new variable into an indefinite integral to produce a function of that variable
2. Differentiate that function using the Leibniz rule
3. Solve the resulting differential equation to get the solution to the integral
So you see it's only 1/3 the Leibniz rule...
Does Feynman work for indefinite integrals?
It doesn't make sense to do this because the indefinite integral is not a function, so there is nothing to differentiate. So no, you can't directly apply Feynman's trick.
I had exactly the same integral in the final exam 1 months ago😂😂
Does n have to be an integer? Seems like any positive real (or possibly any complex number with positive real component) would work.
None of the methods he used assumed n was an integer, so it holds for any positive real, and possibly other complex numbers.
@@rockinroggenrola7277as long as the integral exists
@@jkid1134 It exists for any n>0.
Perfect!👌
Why do we take the partial derivative with respect to y, and why can't we simply take the partial derivative with respect to 'n'?
Bcs n is restricted to natural numbers. You cant differentiate that
Does the American spelling of 'favourite' make anyone else have a mild stroke?
No
No
No
No. Unlike you, we have rejected the spelling of our colonizers. Pardon, colonisateurs. There's that "s" for you. Wouldn't want you to have a stroke.
No
why you keep reposting your old videos like this one ????
Frulanni did this long before Feynman. Also not hard if you expand e^(n*lnx) as a series.
Feynman didn't invent it, he mentions that he saw it in some textbook. It just became popular because he was famous and sociable and probably spread it to others, and since it didn't have an existing name that one stuck
✌
This trick doesn’t belong to Feynman. He learned it from another calculus book, and just popularised it. He didn’t invent it
Of course he didn't invent it. It's just the application of the Leibniz rule. But it's popularly named after Feynman precisely because of what you just wrote
At least he didn't buy it like L'Hipotal.
In one of his popular books, Feynman cited Fredrick S. Woods "Advanced Caculus." Feynman says his high school math teacher gave it to him. I have my dad's copy, from when he tautght at RPI in the late 50's. Sure enough, it's on page 141. Integrating under the integral is on page 145. (1934 edition.)
@@warrickdawes7900 L'Hipotal bought it from whom ?
@@charleyhoward4594 Bernoulli