WARNING: At 2:40, he replaces a function (e^xlnx) with its taylor series. This only works because the taylor series of e^x converges everywhere, so e^f(x) converges everywhere for all functions f(x). However, if you try to do this with a function whose interval of convergence does not cover the interval of integration, then you end up integrating with the a different function outside the interval of convergence than what you started with
@@AH-jt6wcthe Taylor series expansion will converge to the values of exp(x) at any given point. Convergence here is not being used with end behavior in mind. There are functions (example: 1/(1+x^2)) that are not analytic and will not converge to the function it is trying to approximate beyond a certain radius. For this particular example, that radius will be 1 given that you use an expansion at the origin, because the series expansion tries to approximate it for all values including the imaginary unit i, and it doesn’t do well with irregularities like dividing by zero. exp(x) doesn’t have this issue so the Taylor series expansion will converge to exp(x) rather than being limited by something like the previous example.
TYPO 1: I got a bit silly with whether my index variable is called i or n. It doesn't matter, it just needs to be the same throughout! TYPO 2: At 6:30 I use a power rule (a^b)^c=a^(bc). However, I should have make it explicit that there is brackets around all of e^(-u). Clarification: The key trick in solving this integral was writing the integrand as a power series and then interchanging the integral and the summation. If the sum of the integral of the absolute value of a sequence of functions converges, then we can do this interchange (this is a consequence of Fubini's Theorem). In our case, the integrands are either all positive or all negative, depending on n, so taking absolute values is equivalent to removing the negative signs in the resulting series which we can show converges via the ratio test. This justifies the interchange.
You also made use of the fact that the integral over the product measure of a discrete counting measure and a given measure is the same as summing outside of the integral.
At 5:41 when you plug in, you should put e^(-u) in a big pair of parentheses, to emphasize that you execute this exponentiation first, and then raise the result to the power (n+1). The way you wrote it, by convention, means that e is raised to the power (-u)^(n+1), and that is totally different.
Back when I was a math major, we tackled this very integral in one of my advanced calculus courses. After I left college I went into programming. Sadly, I never used much of the math again. So forgive me 40 years of eroded skills. I still love this stuff. Having a strong math backgroud has always served me well.
If u look up cambridge step 3, 2009 question 8 it is on this exact integral and its a very excellent question i remember doing it when studying. Very fun to figure out and pleasing result :)
Another clever way to evaluate the integral from 0 to 1 of x^n*(ln(x))^n w.r.t. x is to use Feynman's trick, i.e. differentiation under the integral sign, a special case of the Leibniz integral rule. Start with the integral I(t), defined as the integral from 0 to 1 of x^t w.r.t. x, with t being a nonnegative continuous variable. I(t) is the first and easiest integral anyone learns in calculus, evaluating to 1/(t+1) after applying the reverse power rule and the Fundamental theorem of calculus. Next notice that, upon differentiation w.r.t. t and exchanging the integral and derivative operators (the derivative becoming a partial derivative under the integral sign), that I'(t) is the integral from 0 to 1 of x^t*ln(x) w.r.t. x. So differentiating I(t) once chains out a single factor of ln(x). This pattern continues, meaning I''(t) is the integral from 0 to 1 of x^t*(ln(x))^2 w.r.t. x, I'''(t) is the integral from 0 to 1 of x^t*(ln(x))^3 w.r.t. x, and so on, since d/dt(x^t) = x^t*ln(x) and ln(x) is a constant with respect to t. Therefore, differentiating I(t) n times and evaluating at t = n returns the integral in question, so I^(n)(n) is the integral from 0 to 1 of x^n*(ln(x))^n w.r.t. x. To evaluate I^(n)(t), look back at I(t). Since I(t) = 1/(t+1), I^(n)(t) = d^n/dt^n(1/(t+1)). Now a seemingly difficult problem of integration has been transformed into a rather simple problem of differentiation. To evaluate this nth derivative, look at the derivative for the first few values of n and try to recognize a pattern. I^(0)(t) = 1/(t+1) = (-1)^0*0!/(t+1)^(0+1), I^(1)(t) = -1/(t+1)^2 = (-1)^1*1!*1/(t+1)^(1+1), I^(2)(t) = 2/(t+1)^3 = (-1)^2*2!*1/(t+1)^(2+1), I^(3)(t) = -6/(t+1)^4 = (-1)^3*3!*1/(t+1)^(3+1)... , the pattern continuing on up to I^(n)(t), suggesting that I^(n)(t) = (-1)^n*n!/(t+1)^(n+1), which can be proved by induction. Finally, evaluating I^(n)(t) at t = n gives the desired result: I^(n)(n) = (-1)^n*n!/(n+1)^(n+1). Some lessons to take away from this solution are 1) that recognizing and exploiting patterns can be very helpful when solving math problems and 2) starting with a similar-looking problem that you know how to solve can often provide some insight into the solution of a more difficult problem. As an aside, this method of differentiation under the integral sign can equivalently be used to prove that Gamma(n+1) = n!, giving a greater intution as to why this technique gives the same result as that of this video, but I will leave that up to you!
@@OtherTheDave Yes that works for finite sums, but infinite sums are defined as the limit of a series. So when you want to interchange them with an integral, you have to interchange a limit and an integral, which isn't always possible.
From a math standpoint, it's fine to do the integral switcheroo thing without justification, so long as you present your findings as a mere conjecture and not as a proof. So in some sense, there's nothing wrong with physicists omitting the justification. They just have to understand that what they've done is found a plausible answer to the question they're asking, but without a proof of the correctness of that answer. Unfortunately, a lot of physics types don't really get and/or worry about these kinds of "subtle" distinctions, despite that they're obviously important, and actually not very subtle at all.
Thanks for this beautiful solution to the Bernoulli integral ∫₀¹xˣdx. Just a small point about the value of 0⁰. At the start of the video you mentioned the apparent contradiction between the rules x⁰=1 and 0ˣ=0, and wondered which rule should take priority when x=0. As far as I am concerned, the x⁰=1 rule definitely takes priority. This is because it is valid for all real x≠0, whilst the 0ˣ=0 rule, on the other hand, only applies for x>0, with 0ˣ being undefined for x0 it is 0, for x0. If we let a→0 and take the pointwise limit of these functions, we get 0 for x>0, 1 for x=0 and ∞ for x
0^7 is zero multiplied seven times. Zero multiplied by anything is zero. x^0 is x multiplied zero times. If you assume an implicit coefficient of 1, then you get 1 multiplied by nothing (not even zero), which is still 1. 0^0 is zero multiplied zero times. Again, assuming an implicit coefficient of 1, then you get 1, because that implicit 1 was never multiplied by zero (there are none).
@@charetjc Agreed. The empty product argument is an excellent justification for 0⁰=1. It is the basis of BriTheMathGuy's video "The Most Controversial Number in Math" (th-cam.com/video/jNhjB4UfR9A/w-d-xo.html ). Another excellent argument (but probably too abstract to be understood by non-Mathematicians) is the map counting argument: if A and B are finite sets with cardinality m and n respectively, then the number of maps A→B (essentially, the number of m-tuples with values in B), written Bᴬ, has cardinality bᵃ. This rule is certainly true if A and B are not both empty, so it is logical to extend it to that case also. There is exactly one map ∅→∅, and that is the empty map ∅. So we get 0⁰=1.
And on top of that, when dealing with integrals we're really only dealing with the open set of values within the boundaries of the integral. And since x^x approaches 1 as x approaches 0 this is the value we're after, not the value at x^x when x=0, regardless of what it is.
Numberphile I think did a nice video about 0^0 and while this intuition works for real numbers, for complex values, it is considerably more... complex. Many of the limits introduced don't converge.
@@crosseyedcat1183 Sure. But the thing is that there are several good reasons to define 0⁰ as a number that have nothing to do with limits. As a limit, 0⁰ is indeterminate, both for real and complex numbers.
Video: Ahh yes. Bernoullis contribution to math. Me: The guy who did the triangles. Nice. Video: Sorry no. The Bernoullis are to math what the Mathews family is to football. Different one
What might be fun is a short video discussing what I was taught to call "Indeterminate forms" such as your example - zero to the power of zero - but including: - anything divided by zero, - infinity (or at least the cardinality of the continuum) to the power of zero, - anything to the power of infinity, and so forth, with attention to the way some make even less sense than others.
Sorry if I'm the 80th person to point this out, but your final summation is indexed by i while the terms are indexed by n. It was correct when you first introduced the Maclaurin series for the exponential function but, after you computed the integrals and put the whole series back together to finish the problem, that was when the i showed up.
This beautiful solution by Johann Bernoulli is justly shown here by the professor, involving e/ln, integration of x, lnx b/w 1 and 0; IBP was not used then, the same Johann who was stumped with Basel problem, and got to admire Euler's solution, (sadly w/o Jakob). You are a teacher I am looking for.
Hello everyone! I might be kinda late, but here's my generalized approach for limits from a to b, being both real numbers. When Dr.Treffor said right before trying to get the gamma function for the answer that you can integrate by parts x^n*ln(x)^n, you can and the formula for the series that will appear is actually easy but a headache to get: Σ (0=k to n) of (x^(n+1)ln(x)^(n-k)*(-1)^k*n!)/((n+1)^(k+1)*(n-k)!). You guys can plug in the values for the summatory and will see how they work. Finally, we can use the Taylor series of e^x to get a generalized integral which can be evaluated from a to b. We get Σ (n=0 to inf) of (1/(n!)*(the aforementioned summatory)). Then just use Barrow's law. Despite it is theoretically correct and perfectly usable, for a computer to evaluate it the summatory will have an error because, obviously, the limit will be a real number and not infinity, not to mention evaluating it manually. Thus, I'd like that the ones that have an extended knowledge in the field can tell me whether it is plausible for the formula to be simplified in some way or method.
The sum at the end is basically 1/x^x for odd x - 1/x^x for even x. For even x this sums to pi^2/6 as in the Basel problem. But no one has been able to find the 1/x^x odd ones yet, hence why he doesn’t show the exact value of convergence
When he said "i will leave this as an exercise for you" i got severe PTSD of my maths professors at uni. Very rarely did they show us how results came about lol
@@ChemicaLove what is intuitively obvious for a college professor who has years of research under their belt may not be the case for students. I dont feel like leaving properties or theorems for students to find on their own is particularly useful when the work is relatively new to the students. Feel like it rarely had the desired effect and instead led to a lot of acceptance of mathematical facts without knowing why, a teaching practice which is holistically frowned upon in mathematics
@@badnoodlez what an absolutely gross misinterpretation of what I said. You don't think students have enough work to do without being left to piece together theorems and properties from work that is relatively new to them (the entire concept of 'proofs' was a very new thing to myself in my maths degree)? Challenging students to discover for themselves is a great tool when used correctly, but was totally abused by professors during my own degree experience. It came across that they clearly prioritised their other work and research over lecturing and used this "leave this to you as an exercise" as a mere shortcut so that they didn't have to bother spending any more time elaborating. Most of what they asked us to prove on our own as students we were not capable of and they rarely provided solutions for those who were unsure. Myself and many others had to resort to online tutorials and lectures over our own professors.
To finish it off split the series into the odd and even parts to get rid of the (-1)^n, then re index the two sums to get the Basel problem and plug in pi^2/6
I admit I like to think about the way interections within the function suggest limitations on them. For example consider this, the existence of the power rule for derivatives seems to suggest that for whole number all odd slope that are on a x of odd power will have come from a fraction and that for a whole number c that could represent any number, there is potentiolly a limit on how low it can actually be if all numbers are whole due to how power will keep multiplying up. I have a 44 page paper on the collatz conjecture, but my record of passing math classes is admittily not perfect due to several issues with hand writing and expression of knowledge on tests themselves... your channel is always a good reminder for fun. Sometimes difficult to find less known rules that you discover yourself but that makes it interesting until you want to confirm they exist already lol XD
x^x is such a bad boy. You've gotta be a math nerd to understand, but I just like that function. I am curious however whether it actually represents something in the real world, in the same way that n! expresses the number of permutations of n objects.
How many possible mappings are there of a set of cardinality n with itself? Each element can be mapped to any of n possible elements, since there are n elements, we have n times n times n... n times, so n^n. Combinatorially, a^n is more versatile. If I have a 4 digit pin, there are 10^4 pins I could choose from, since for the first number there are 10 options, for the second 10 options, same for third and fourth. Similar principle. This kind of discrete math appears a lot in applications.
Well we could make it even more cleaner by assuming that (-1)^n =-(-1)^(n+1) than make a shift by one in the serie expression to fin the resulat equal to the sum from 1 to infinity of -(-1/n)^n
i once actually did the n-fold integration by parts cus i thought it was interesting. honestly im surprised i haven't seen anyone cover that method in a video
5:44 The brackets should enclose the entire e-function, not only the exponent. There are already a couple of other comments wondering how you simplified further using the power rule, which was correct, but didn't make sense if the brackets are only about the -u in the exponent.
I was thinking the same, then I rewatched the video and saw that it come from x^n --> (e^-u)^(n+1), so it's correct what he did. It's for the notation that it seems wrong. Anyway, very good video
This, along with an even nicer integral, is collectively known as the "Sophomore's Dream"! The other integrand is x^-x and the result is even nicer: integral(x^-x) = sum(n^-n)
With the first graph, if you plug it in in Desmos, even though desmos doesnt understand imaginary numbers, when you click home it sometimes shows things such as -1/3 as a value, although you cant click on it because buggy desmos
clicking on your video i already figured out that 0^0 = 1 from knowing the behavior of root n of x function tend to grow toward a step function at n=infinity (power of 0). and 1^1 = 1.
Eeeeh waitaminute. @8:40 The condition of absolute convergence is not a thing (it has other properties, though, such as the re-ordering of the terms). In order to swap the integral and the summation signs, you need either: 1) uniform convergence; this is the case, because the exponential series Σ y^n /n! has a infinite radius, therefore UC, and even normal convergence is guaranted on [-1,0] or whatever interval (x ln x) spans; 2) dominated convergence; on the interval [0,1], | x^x | = | exp(x lnx) | < 1 and ∫ 1 dx is well defined on [0..1] so it's ok; 3) monotonous convergence; (x lnx)^n / n! < 0 so the series is strictly decreasing as n goes to ∞, so it's ok. Forgive me if I'm completely mistaken, because it's 1 AM, I just came back from a festival, I'm exhausted and I'm doing math, ah ah.
At 7:44 -- I am sure it was just an oversight, the equation evaluates correctly replacing the i with n, but I was looking to see if anyone else noticed it.
I remember discussing difficult to integrate functions in my undergraduate computer science numerical methods class. And I remember thinking the Newton was genius to see that infinite trapezoids could be used to solve the problem. I also felt badly that Newton lived in a time before computers, but thank goodness I was alive NOW! 🙂
3:03 The taylor series for e^(xlnx) cant just be expanded like that from e^x xlnx is undefined for zero, so the best expansion we can get is e^xlnx=sum of (xlnx-1)^n/n!
It’s legal to make a substitution that turns a proper integral into an improper integral (e.g. by making the integrand undefined at one endpoint of the interval) as long as the relevant limits exist when we need to take limits.
@@bfrobin446 well the limit of lnx is -inf as x approaches 0, so e^xlnx is undefined . I am still confused and not sure how you can write this expansion as that
@@Alex-xu3mo e^(x ln x) is defined on the entire integration interval except at x=0, and the Taylor series that’s shown is correct everywhere the original expression is defined. So we make the substitution and then implicitly treat the integral as an improper integral: instead of integrating from 0 to 1, integrate from ε to 1 and then take the limit as epsilon goes to zero from the right.
@@bfrobin446 @Ben Robinson i dont understand how that particular substitution can be made. We have the function e^xlnx which's first derivative is e^xlnx(lnx+1) and the following ones have more lnx+1 and 1/x in them. I dont get how we can build a taylor series from that so that any derivative at particular value in the series becomes 1, and we get neat expression xlnx^n/n! Precisely i dont get how ((e^(alna))^(n)*(xlnx-a)^n)/n! taylor series n member can become ((xlnx)^n)/n!
When you switched the integral with the sum in the beginning even before you said it I was thinking dominated convergence theorem or monotone/bounded CT
When taking eˆxln(x). Since we are working on an integral from 0 to 1, x can be egal to 0 and so ln(0) isn't define. Have we still the right to use ln in this case? Like isn't replacing xˆx by eˆxln(x) impossible since the 2nd expression isn't defined in 0?
I don't understand your argument for switching the order of integration and summation. I can see it with the dominated convergence theorem using sup |x^n * ln(x)^n| = 1/e^n on [0,1] (1/(1-1/e) will do as majorant because of geometric series) but I don't see how the fact that the series of the already evaluated integrals is absolutely convergent yields this result.
Let me elaborate. I’m citing without stating it a consequence of Fubini’s theorem that if the sum of the integral of |f_n| converges, then you can interchange the sum and integral without absolute values. In our case the absolute value can be taken before or after the integral because the f_n is either all negative or all positive on our interval of study in this example, but you are absolutely right that it isn’t generally true to do it with the absolute value on the outside of the integral.
the justification of the switcheroo was a circular argument. you state that the switcheroo is justified if it converges, but you then got a converging series by performing a switcheroo.
This was a fun one. Made a good break after my finance class. Question, do you think a Fuchsian group can be rendered in 3-dimensional hyperbolic space? Since hyperbolic space is weird, and Fuchsian groups are just projections into the hyperbolic space. (Sorry, doing research on Fuchsian groups and the question has been niggling at my brain for a minute.)
Thank you for this journey down memory lane. Once we realized that Calculus is Algebra with limits, mostly Algebra, it became much easier. Just remember that the endpoints of an integral are not included in the value, because it's ultimately a limit operation. Using a program to evaluate something like 0^0 is just accepting a definition, they simply define 0^0 to be 1. Computers are detrimental to understanding.
I would still ask for proof of speed of convergence. What if converges really slowly and terms appear to stabilize only to change a lot later on in the expansion?
It's interesting but there is a simple reason! Taking the derivative of x, we find it to be ln x+1(x^x). Now if we set it to 0 and since here 0^0=1, we see (ln x+1)(x^x) is attained when ln x=-1, so x=e^(-1)=1/e. We can verify it is a minimum by checking if f''(1/e)>0: d((ln x+1)(x^x))/dx=(ln x+1)(ln x+1)(x^x)+(1/x)(x^x)={x^(x-1)}{(ln x+1)^2+1} (feel free to correct errors!) We see this as a local minimum as when x=1/e, it turns out to be (1/e)^((1-e)/e), which is positive.
x = x * 1 x^y = x^y * 1 x^0 = 1 (there are no x's to raise to y) Therefore... 0^0 = 1 The only real debate to be had is whether -0^0 and/or 0^-0 are 1 or -1
Thanks for this showing this beautiful piece of mathematics btw I am so tiny to praise you sir 😅but I have a question which I want to ask Which shape (2d,3d) covers most area for least perimeter. I think it should be circle for regular shapes since it has highest area perimeter ratio but don't know if any complex shape in 3D space ( some sort of bent circular shape, like Pringles or anything) does so.
@amr m I think they meant higher dimensional versions of perimeter, for 3d it would be surface area, for 4d it would be volume, etc Either way, the correct answer is a sphere. In fact, I believe spheres get more efficient in higher and higher dimensions
because the power rule for antiderivatives only works if the exponent is a constant, which it isn't in this case the antiderivative of x^x is not x^(x+1) / (x+1), if you try to take the derivative of x^(x+1) / (x+1) you actually get x^x * (x^2 * ln(x) + x * ln(x) + x^2 + x + 1) / (x+1)^2 in fact, x^x does not have a nice elementary antiderivative at all
WARNING: At 2:40, he replaces a function (e^xlnx) with its taylor series. This only works because the taylor series of e^x converges everywhere, so e^f(x) converges everywhere for all functions f(x). However, if you try to do this with a function whose interval of convergence does not cover the interval of integration, then you end up integrating with the a different function outside the interval of convergence than what you started with
the exp(x) function diverge when x tend to infinity right ? you said converge everywhere
@@AH-jt6wcthe Taylor series expansion will converge to the values of exp(x) at any given point. Convergence here is not being used with end behavior in mind.
There are functions (example: 1/(1+x^2)) that are not analytic and will not converge to the function it is trying to approximate beyond a certain radius. For this particular example, that radius will be 1 given that you use an expansion at the origin, because the series expansion tries to approximate it for all values including the imaginary unit i, and it doesn’t do well with irregularities like dividing by zero.
exp(x) doesn’t have this issue so the Taylor series expansion will converge to exp(x) rather than being limited by something like the previous example.
smartest koisheep
Yes for definite integral, when you do substitution, you need to change the range?
damm koishi
this professor never lets me stay away from mathematics.
math will grip you forever! :D
@@DrTrefor The limit of y=x^x as x approaches 0 is undefined. Doesn't that mean that the integral in your video is undefined?
@@paulthompson9668 plus you only really need it to exist almost everywhere to integrate so having the endpoint undefined doesnt matter
@@cnutsiggardason2014 Is that a rule for integration? I'm not trying to be pedantic, but sometimes a single point matters and sometimes it doesn't.
He will find you and he will math you. There is no escape. Even TH-cam is not safe.
The Bernoullis reproduced like rabbits. Should have better studied Fibonacci kind of problems
Goddamn intelligence runs in the family
The Fibonacci sequence grows in an exponential order so maybe that's not such a great idea either
@@darkking2436 Makes a 69 sort of pattern, so less kids, more fun
Okay. That was good.
I would have hoped they would have involved more people outside the family when reproducing, than Fibonacci's rabbits did.
TYPO 1: I got a bit silly with whether my index variable is called i or n. It doesn't matter, it just needs to be the same throughout!
TYPO 2: At 6:30 I use a power rule (a^b)^c=a^(bc). However, I should have make it explicit that there is brackets around all of e^(-u).
Clarification: The key trick in solving this integral was writing the integrand as a power series and then interchanging the integral and the summation. If the sum of the integral of the absolute value of a sequence of functions converges, then we can do this interchange (this is a consequence of Fubini's Theorem). In our case, the integrands are either all positive or all negative, depending on n, so taking absolute values is equivalent to removing the negative signs in the resulting series which we can show converges via the ratio test. This justifies the interchange.
Integral of the cube root of tan(x)?
thanks, I was really confused by the second typo.
@@stackalloc7741 me too! I came to the comments to ask precisely that and see it was already clarified.
You also made use of the fact that the integral over the product measure of a discrete counting measure and a given measure is the same as summing outside of the integral.
I be confusion in this equation, but know I understood
At 5:41 when you plug in, you should put e^(-u) in a big pair of parentheses, to emphasize that you execute this exponentiation first, and then raise the result to the power (n+1). The way you wrote it, by convention, means that e is raised to the power (-u)^(n+1), and that is totally different.
I was going to write the exact same comment on the exponent. Still good videos though
Thanks for pointing out. I was also confused how this could be equal. The quote position is actually not correct.
@@amrm155that's like remote punch right into the stomach.
Thanks for pointing this out. I was having the same issue as other people and thought he was doing some ninja math.
@amr m Don't you always write i=0 under the sigma
Back when I was a math major, we tackled this very integral in one of my advanced calculus courses. After I left college I went into programming. Sadly, I never used much of the math again. So forgive me 40 years of eroded skills. I still love this stuff. Having a strong math backgroud has always served me well.
Your presentation of this problem is really motivating me to try on my own such incredible integrals !
If u look up cambridge step 3, 2009 question 8 it is on this exact integral and its a very excellent question i remember doing it when studying. Very fun to figure out and pleasing result :)
Another clever way to evaluate the integral from 0 to 1 of x^n*(ln(x))^n w.r.t. x is to use Feynman's trick, i.e. differentiation under the integral sign, a special case of the Leibniz integral rule. Start with the integral I(t), defined as the integral from 0 to 1 of x^t w.r.t. x, with t being a nonnegative continuous variable. I(t) is the first and easiest integral anyone learns in calculus, evaluating to 1/(t+1) after applying the reverse power rule and the Fundamental theorem of calculus. Next notice that, upon differentiation w.r.t. t and exchanging the integral and derivative operators (the derivative becoming a partial derivative under the integral sign), that I'(t) is the integral from 0 to 1 of x^t*ln(x) w.r.t. x. So differentiating I(t) once chains out a single factor of ln(x). This pattern continues, meaning I''(t) is the integral from 0 to 1 of x^t*(ln(x))^2 w.r.t. x, I'''(t) is the integral from 0 to 1 of x^t*(ln(x))^3 w.r.t. x, and so on, since d/dt(x^t) = x^t*ln(x) and ln(x) is a constant with respect to t. Therefore, differentiating I(t) n times and evaluating at t = n returns the integral in question, so I^(n)(n) is the integral from 0 to 1 of x^n*(ln(x))^n w.r.t. x. To evaluate I^(n)(t), look back at I(t). Since I(t) = 1/(t+1), I^(n)(t) = d^n/dt^n(1/(t+1)). Now a seemingly difficult problem of integration has been transformed into a rather simple problem of differentiation. To evaluate this nth derivative, look at the derivative for the first few values of n and try to recognize a pattern. I^(0)(t) = 1/(t+1) = (-1)^0*0!/(t+1)^(0+1), I^(1)(t) = -1/(t+1)^2 = (-1)^1*1!*1/(t+1)^(1+1), I^(2)(t) = 2/(t+1)^3 = (-1)^2*2!*1/(t+1)^(2+1), I^(3)(t) = -6/(t+1)^4 = (-1)^3*3!*1/(t+1)^(3+1)... , the pattern continuing on up to I^(n)(t), suggesting that I^(n)(t) = (-1)^n*n!/(t+1)^(n+1), which can be proved by induction. Finally, evaluating I^(n)(t) at t = n gives the desired result: I^(n)(n) = (-1)^n*n!/(n+1)^(n+1). Some lessons to take away from this solution are 1) that recognizing and exploiting patterns can be very helpful when solving math problems and 2) starting with a similar-looking problem that you know how to solve can often provide some insight into the solution of a more difficult problem. As an aside, this method of differentiation under the integral sign can equivalently be used to prove that Gamma(n+1) = n!, giving a greater intution as to why this technique gives the same result as that of this video, but I will leave that up to you!
This is basically the same as iterating integration by parts
@@TheEternalVortex42 How? It seems to me that, although both solutions give the same answer, they are fundamentally different.
I always forget you can’t always swap sums and integrals. My physics profs always just do it without comment
But you shouldn't tell that to mathematicians! ;)
The nice thing about physics is that most things in the world are sufficiently nonpathological such that you can play fast and loose pretty safely 😅
Wait, why not? I was always taught that “the sum of the integrals is the integral of the sum”.
@@OtherTheDave Yes that works for finite sums, but infinite sums are defined as the limit of a series. So when you want to interchange them with an integral, you have to interchange a limit and an integral, which isn't always possible.
From a math standpoint, it's fine to do the integral switcheroo thing without justification, so long as you present your findings as a mere conjecture and not as a proof. So in some sense, there's nothing wrong with physicists omitting the justification. They just have to understand that what they've done is found a plausible answer to the question they're asking, but without a proof of the correctness of that answer. Unfortunately, a lot of physics types don't really get and/or worry about these kinds of "subtle" distinctions, despite that they're obviously important, and actually not very subtle at all.
Never stop makin videos professor u are an absolute gem of a person
Thanks for this beautiful solution to the Bernoulli integral ∫₀¹xˣdx.
Just a small point about the value of 0⁰. At the start of the video you mentioned the apparent contradiction between the rules x⁰=1 and 0ˣ=0, and wondered which rule should take priority when x=0.
As far as I am concerned, the x⁰=1 rule definitely takes priority.
This is because it is valid for all real x≠0, whilst the 0ˣ=0 rule, on the other hand, only applies for x>0, with 0ˣ being undefined for x0 it is 0, for x0. If we let a→0 and take the pointwise limit of these functions, we get 0 for x>0, 1 for x=0 and ∞ for x
0^7 is zero multiplied seven times. Zero multiplied by anything is zero.
x^0 is x multiplied zero times. If you assume an implicit coefficient of 1, then you get 1 multiplied by nothing (not even zero), which is still 1.
0^0 is zero multiplied zero times. Again, assuming an implicit coefficient of 1, then you get 1, because that implicit 1 was never multiplied by zero (there are none).
@@charetjc Agreed. The empty product argument is an excellent justification for 0⁰=1. It is the basis of BriTheMathGuy's video "The Most Controversial Number in Math" (th-cam.com/video/jNhjB4UfR9A/w-d-xo.html ).
Another excellent argument (but probably too abstract to be understood by non-Mathematicians) is the map counting argument: if A and B are finite sets with cardinality m and n respectively, then the number of maps A→B (essentially, the number of m-tuples with values in B), written Bᴬ, has cardinality bᵃ. This rule is certainly true if A and B are not both empty, so it is logical to extend it to that case also. There is exactly one map ∅→∅, and that is the empty map ∅. So we get 0⁰=1.
And on top of that, when dealing with integrals we're really only dealing with the open set of values within the boundaries of the integral. And since x^x approaches 1 as x approaches 0 this is the value we're after, not the value at x^x when x=0, regardless of what it is.
Numberphile I think did a nice video about 0^0 and while this intuition works for real numbers, for complex values, it is considerably more... complex. Many of the limits introduced don't converge.
@@crosseyedcat1183 Sure. But the thing is that there are several good reasons to define 0⁰ as a number that have nothing to do with limits. As a limit, 0⁰ is indeterminate, both for real and complex numbers.
Absolutely loved this video. Thank you Trefor for bringing this gorgeous integral ^^
Glad you enjoyed it!
Great example and explanation professor!
Video: Ahh yes. Bernoullis contribution to math.
Me: The guy who did the triangles. Nice.
Video: Sorry no. The Bernoullis are to math what the Mathews family is to football. Different one
What might be fun is a short video discussing what I was taught to call "Indeterminate forms" such as your example - zero to the power of zero - but including:
- anything divided by zero,
- infinity (or at least the cardinality of the continuum) to the power of zero,
- anything to the power of infinity,
and so forth, with attention to the way some make even less sense than others.
すばらしい。どうしても解けなかった問題を解いてくれた。
Sorry if I'm the 80th person to point this out, but your final summation is indexed by i while the terms are indexed by n. It was correct when you first introduced the Maclaurin series for the exponential function but, after you computed the integrals and put the whole series back together to finish the problem, that was when the i showed up.
Excellent presentation, professor. ⭐️
This beautiful solution by Johann Bernoulli is justly shown here by the professor, involving e/ln, integration of x, lnx b/w 1 and 0; IBP was not used then, the same Johann who was stumped with Basel problem, and got to admire Euler's solution, (sadly w/o Jakob). You are a teacher I am looking for.
Great video thanks,
In the challenge examples at the end, the first one should be :
Int 0 to 1 (1/(x^x)) dx = sum (n=1 to inf) 1/(n^n) and not 1/(n^2)
Hello everyone! I might be kinda late, but here's my generalized approach for limits from a to b, being both real numbers. When Dr.Treffor said right before trying to get the gamma function for the answer that you can integrate by parts x^n*ln(x)^n, you can and the formula for the series that will appear is actually easy but a headache to get: Σ (0=k to n) of (x^(n+1)ln(x)^(n-k)*(-1)^k*n!)/((n+1)^(k+1)*(n-k)!). You guys can plug in the values for the summatory and will see how they work. Finally, we can use the Taylor series of e^x to get a generalized integral which can be evaluated from a to b. We get Σ (n=0 to inf) of (1/(n!)*(the aforementioned summatory)). Then just use Barrow's law.
Despite it is theoretically correct and perfectly usable, for a computer to evaluate it the summatory will have an error because, obviously, the limit will be a real number and not infinity, not to mention evaluating it manually. Thus, I'd like that the ones that have an extended knowledge in the field can tell me whether it is plausible for the formula to be simplified in some way or method.
The sum at the end is basically
1/x^x for odd x - 1/x^x for even x. For even x this sums to pi^2/6 as in the Basel problem. But no one has been able to find the 1/x^x odd ones yet, hence why he doesn’t show the exact value of convergence
Really trending video, great job!
that's one great piece of math right there!! thank you
Glad you think so!
@Dr. Trefor Bazett I really learned a lot from you and am still learning. I appreciate your great work.
When he said "i will leave this as an exercise for you" i got severe PTSD of my maths professors at uni. Very rarely did they show us how results came about lol
The answer is intuitively obvious and left as an exercise for the student..
@@ChemicaLove what is intuitively obvious for a college professor who has years of research under their belt may not be the case for students. I dont feel like leaving properties or theorems for students to find on their own is particularly useful when the work is relatively new to the students. Feel like it rarely had the desired effect and instead led to a lot of acceptance of mathematical facts without knowing why, a teaching practice which is holistically frowned upon in mathematics
@@steverainingagain7956so you don't think students should work and obviously the prof spent their PhD ruminating over trivial exercises.
😂😂😂😂
@@badnoodlez what an absolutely gross misinterpretation of what I said. You don't think students have enough work to do without being left to piece together theorems and properties from work that is relatively new to them (the entire concept of 'proofs' was a very new thing to myself in my maths degree)? Challenging students to discover for themselves is a great tool when used correctly, but was totally abused by professors during my own degree experience. It came across that they clearly prioritised their other work and research over lecturing and used this "leave this to you as an exercise" as a mere shortcut so that they didn't have to bother spending any more time elaborating. Most of what they asked us to prove on our own as students we were not capable of and they rarely provided solutions for those who were unsure. Myself and many others had to resort to online tutorials and lectures over our own professors.
Oh neat, the tetrate x^^2.
Wonderfully done. I throughly enjoyed your process and pacing.
0^0 is obviously 1 since it is the cardinality of the set of functions which map the empty set to itself, namely the empty function an no others
To finish it off split the series into the odd and even parts to get rid of the (-1)^n, then re index the two sums to get the Basel problem and plug in pi^2/6
No, that isn't the same formula, the basel problem is x^(-2) not x^(-x)
I admit I like to think about the way interections within the function suggest limitations on them. For example consider this, the existence of the power rule for derivatives seems to suggest that for whole number all odd slope that are on a x of odd power will have come from a fraction and that for a whole number c that could represent any number, there is potentiolly a limit on how low it can actually be if all numbers are whole due to how power will keep multiplying up. I have a 44 page paper on the collatz conjecture, but my record of passing math classes is admittily not perfect due to several issues with hand writing and expression of knowledge on tests themselves... your channel is always a good reminder for fun. Sometimes difficult to find less known rules that you discover yourself but that makes it interesting until you want to confirm they exist already lol XD
I’m 73 and my math knowledge stays at the level of sophomore in college.
Still I find your topics are always interesting and intriguing.😊😊😊
I love that you are still learning!
integral from 0 to 1 of x^x dx = - sum of n from 1 to infinity of (-1)^n * n^-n
Professor you blow my mind
in 9:31 the "challenge examples" in the first integral of 1/x^x the result is not the sum of 1/n^2 it's the sum of 1/n^n
x^x is such a bad boy. You've gotta be a math nerd to understand, but I just like that function. I am curious however whether it actually represents something in the real world, in the same way that n! expresses the number of permutations of n objects.
How many possible mappings are there of a set of cardinality n with itself? Each element can be mapped to any of n possible elements, since there are n elements, we have n times n times n... n times, so n^n. Combinatorially, a^n is more versatile. If I have a 4 digit pin, there are 10^4 pins I could choose from, since for the first number there are 10 options, for the second 10 options, same for third and fourth. Similar principle. This kind of discrete math appears a lot in applications.
n^n occasionally came up when calculating entropy, but I’m not a thermodynamist, so I can’t tell you how often it useful it is
Using the Basal problem and the steps to solve it, you can get a final solution to this as n goes to infinity of pi^2/12
Well we could make it even more cleaner by assuming that (-1)^n =-(-1)^(n+1) than make a shift by one in the serie expression to fin the resulat equal to the sum from 1 to infinity of -(-1/n)^n
7:36 isn't that pi function since gamma function would be n-1 instead of just n.
i once actually did the n-fold integration by parts cus i thought it was interesting. honestly im surprised i haven't seen anyone cover that method in a video
5:44 The brackets should enclose the entire e-function, not only the exponent. There are already a couple of other comments wondering how you simplified further using the power rule, which was correct, but didn't make sense if the brackets are only about the -u in the exponent.
Yes, I should have been explicit, I'll add a note
Love your content !!
Glad you enjoy!
Lovely integral bud
6:30 it looked like n+1 was the exponent on -u alone so I didn't think you could do that step like that
I was thinking the same, then I rewatched the video and saw that it come from x^n --> (e^-u)^(n+1), so it's correct what he did. It's for the notation that it seems wrong. Anyway, very good video
This, along with an even nicer integral, is collectively known as the "Sophomore's Dream"! The other integrand is x^-x and the result is even nicer:
integral(x^-x) = sum(n^-n)
Typo. 8:09 for example the infinite sum's index shall be n and not i.
With the first graph, if you plug it in in Desmos, even though desmos doesnt understand imaginary numbers, when you click home it sometimes shows things such as -1/3 as a value, although you cant click on it because buggy desmos
Excellent and enjoyable presentation.
great explanations! Could it be that you you messed up the variables i and n since the sum counted the variable i but you used n?
regarding minute 7:38 onwards
Quite right, thank you!
clicking on your video i already figured out that 0^0 = 1 from knowing the behavior of root n of x function tend to grow toward a step function at n=infinity (power of 0). and 1^1 = 1.
Brilliant, beautiful derivation. Thank you
I've got a print of a painting of two brothers Bernoulli deep into some maths hanging above my desk. They wore curly hair wigs like judges used to do.
I Liked too much this solution, I hope more videos
That was pretty creative
Great video, you're an excellent professor :)
Eeeeh waitaminute. @8:40 The condition of absolute convergence is not a thing (it has other properties, though, such as the re-ordering of the terms). In order to swap the integral and the summation signs, you need either:
1) uniform convergence; this is the case, because the exponential series Σ y^n /n! has a infinite radius, therefore UC, and even normal convergence is guaranted on [-1,0] or whatever interval (x ln x) spans;
2) dominated convergence; on the interval [0,1], | x^x | = | exp(x lnx) | < 1 and ∫ 1 dx is well defined on [0..1] so it's ok;
3) monotonous convergence; (x lnx)^n / n! < 0 so the series is strictly decreasing as n goes to ∞, so it's ok.
Forgive me if I'm completely mistaken, because it's 1 AM, I just came back from a festival, I'm exhausted and I'm doing math, ah ah.
Thanks 🙏🏻 teacher . Me from Cambodia 🇰🇭
why did you transform the sum n=0 to n=infinity into sum i=0 to i=infinity at the end without change n from gamma function ?
At 7:44 -- I am sure it was just an oversight, the equation evaluates correctly replacing the i with n, but I was looking to see if anyone else noticed it.
I remember a similar integral being called Sophomore’s Dream, the integrand is x^-x instead of x^x
07:45 The symbol in the summation sign is wrong! It should be "n" instead of "i".
3:25 you called it an improper integral. Is this because of the point at x=0?
Exactly. It isn’t defined at 0, but one can take a limit of that function.
Oh ok thanks
Excellent explanation - great👌
Love the t-shirt!
I remember discussing difficult to integrate functions in my undergraduate computer science numerical methods class. And I remember thinking the Newton was genius to see that infinite trapezoids could be used to solve the problem. I also felt badly that Newton lived in a time before computers, but thank goodness I was alive NOW! 🙂
3:03
The taylor series for e^(xlnx) cant just be expanded like that from e^x
xlnx is undefined for zero, so the best expansion we can get is e^xlnx=sum of (xlnx-1)^n/n!
It’s legal to make a substitution that turns a proper integral into an improper integral (e.g. by making the integrand undefined at one endpoint of the interval) as long as the relevant limits exist when we need to take limits.
@@bfrobin446 well the limit of lnx is -inf as x approaches 0, so e^xlnx is undefined . I am still confused and not sure how you can write this expansion as that
@@Alex-xu3mo e^(x ln x) is defined on the entire integration interval except at x=0, and the Taylor series that’s shown is correct everywhere the original expression is defined. So we make the substitution and then implicitly treat the integral as an improper integral: instead of integrating from 0 to 1, integrate from ε to 1 and then take the limit as epsilon goes to zero from the right.
@@bfrobin446 @Ben Robinson i dont understand how that particular substitution can be made. We have the function e^xlnx which's first derivative is e^xlnx(lnx+1) and the following ones have more lnx+1 and 1/x in them. I dont get how we can build a taylor series from that so that any derivative at particular value in the series becomes 1, and we get neat expression xlnx^n/n!
Precisely i dont get how ((e^(alna))^(n)*(xlnx-a)^n)/n! taylor series n member can become ((xlnx)^n)/n!
@@Alex-xu3mo Let u = x ln x. Write the Taylor series for e^u, then change each u in that series back to an x ln x.
I think that uniform convergence justifies the commutation of series and integral, but correct me if im wrong
Sir Pls make a Video series On Beta and Gamma Function Seperately 🙏.
Good idea! I've done gamma in a few places, but beta function would be a good one.
When you switched the integral with the sum in the beginning even before you said it I was thinking dominated convergence theorem or monotone/bounded CT
i think i need to learn more algebra absolutely love it.
This is more in the lines of analysis.
When taking eˆxln(x). Since we are working on an integral from 0 to 1, x can be egal to 0 and so ln(0) isn't define. Have we still the right to use ln in this case? Like isn't replacing xˆx by eˆxln(x) impossible since the 2nd expression isn't defined in 0?
I don't understand your argument for switching the order of integration and summation. I can see it with the dominated convergence theorem using sup |x^n * ln(x)^n| = 1/e^n on [0,1] (1/(1-1/e) will do as majorant because of geometric series) but I don't see how the fact that the series of the already evaluated integrals is absolutely convergent yields this result.
Let me elaborate. I’m citing without stating it a consequence of Fubini’s theorem that if the sum of the integral of |f_n| converges, then you can interchange the sum and integral without absolute values. In our case the absolute value can be taken before or after the integral because the f_n is either all negative or all positive on our interval of study in this example, but you are absolutely right that it isn’t generally true to do it with the absolute value on the outside of the integral.
Ah, I see. Thank you for elaborating!
This constant is clearly irrational by the rate of convergence. Is it known to be transcendental?
Very good computation ❤
I don't follow the rule of exponents thing at around 6:35 where you bring down the (n+1)...I mean isn't it like saying exp(X**3) = exp(3x)?
I didn't put the brackets in explicitly, but the derivation has it as (e^a)^b not e^(a^b)
@@DrTrefor ah ok, gottit. Power of a power rule Thanks very much 😊. P
You are my inspiration Sir!
Can you make a playlist on series and sequences
Check out my calc II playlist!
the justification of the switcheroo was a circular argument. you state that the switcheroo is justified if it converges, but you then got a converging series by performing a switcheroo.
1:18 You can't you L'Hopital here, because the limits of of numerator and denominator aren't the same, right?
Could you tell me where you purchased that beautiful T-shirt? Plz :)
Interesting! Thanks.
It has some real values for negative xs, so like the gamma function. 😀
when Tom cruise flies on army planes
Understanding this is such a W.
This was a fun one. Made a good break after my finance class. Question, do you think a Fuchsian group can be rendered in 3-dimensional hyperbolic space? Since hyperbolic space is weird, and Fuchsian groups are just projections into the hyperbolic space. (Sorry, doing research on Fuchsian groups and the question has been niggling at my brain for a minute.)
Thank you! I have no idea, I'd have to think about that one:D
Why do you not use Keppler: A = (b-a)/4 * ( f(a) + 2f((a+b)/2) + f(b)) = 0,79
That was pretty crazy.
Great integral
@ 5:45 it shouldn't be e^((-u)^(n+1)), but rather (e^(-u))^(n+1).
Thank you for this journey down memory lane. Once we realized that Calculus is Algebra with limits, mostly Algebra, it became much easier.
Just remember that the endpoints of an integral are not included in the value, because it's ultimately a limit operation.
Using a program to evaluate something like 0^0 is just accepting a definition, they simply define 0^0 to be 1. Computers are detrimental to understanding.
7:57 the index in the bottom of the sigma symbol should read n=0 instead of i=0
I would still ask for proof of speed of convergence. What if converges really slowly and terms appear to stabilize only to change a lot later on in the expansion?
nice video, thank you!
It is also interesting that x^x has a minimum at x=1/e.
It's interesting but there is a simple reason! Taking the derivative of x, we find it to be ln x+1(x^x). Now if we set it to 0 and since here 0^0=1, we see (ln x+1)(x^x) is attained when ln x=-1, so x=e^(-1)=1/e. We can verify it is a minimum by checking if f''(1/e)>0: d((ln x+1)(x^x))/dx=(ln x+1)(ln x+1)(x^x)+(1/x)(x^x)={x^(x-1)}{(ln x+1)^2+1} (feel free to correct errors!)
We see this as a local minimum as when x=1/e, it turns out to be (1/e)^((1-e)/e), which is positive.
@@bengal_tiger1984 I think there is a solution without assuming 0^0=1. y'/y=ln(x) +1=0 => ln(x)=-1 => x=1/e
And y'' = y'^2/y + y/x is always positive.
x = x * 1
x^y = x^y * 1
x^0 = 1 (there are no x's to raise to y)
Therefore...
0^0 = 1
The only real debate to be had is whether -0^0 and/or 0^-0 are 1 or -1
Thanks for this showing this beautiful piece of mathematics btw I am so tiny to praise you sir 😅but I have a question which I want to ask
Which shape (2d,3d) covers most area for least perimeter. I think it should be circle for regular shapes since it has highest area perimeter ratio but don't know if any complex shape in 3D space ( some sort of bent circular shape, like Pringles or anything) does so.
@amr m I think they meant higher dimensional versions of perimeter, for 3d it would be surface area, for 4d it would be volume, etc
Either way, the correct answer is a sphere. In fact, I believe spheres get more efficient in higher and higher dimensions
Sphere is always going to be the best. Technically you can define a sphere as a shape with the highest volume to surface area ratio.
I want to write something in between: "I am fully confused, but also satisfied now." to "pfff... Wasn't that clear in the first place?"
Wait. I’m new to Calculus but why, since this is an integration, is this not
0^1 / 1
minus
1^2/2
to give us 1/2 as the area under the curve?
because the power rule for antiderivatives only works if the exponent is a constant, which it isn't in this case
the antiderivative of x^x is not x^(x+1) / (x+1), if you try to take the derivative of x^(x+1) / (x+1) you actually get x^x * (x^2 * ln(x) + x * ln(x) + x^2 + x + 1) / (x+1)^2
in fact, x^x does not have a nice elementary antiderivative at all
@@mjkhoi6961 Ah, ok. Thank you for the very complete explanation. Much appreciated.
Why can't we take anti derivative? Integral is x^x/(1+lnx) which on limits give 1
Can someone explain why when converting x^n in terms of u, you add 1 to n? In other words why does the exponent become n+1 and not just n?
In the conversion from du to dx there is one more x factor there
Oh I see now, thanks Dr. Bazett!
Is there a closed-form solution to the infinite sum?
I don't believe so!
Is there any important link between this integral and the integral of x TO THE -X FROM 0 TO INFINITY? [Sorry keyboard is broken>}