That looked like a devilishly difficult problem at first. Then you twirled your chalk and, as if by magic, everything fell into place. You are a consummate teacher, young man, and I salute you.
This, I had in high school. Afterwards I had another 2 years of math. So, not yet, but slowly but surely you're getting to topics I have never heard of before and you are starting to loose me. Eventually I will have to stick to math again in order to follow. But it's fun. I just need a little time to keep up. Thank you so much for bringing long forgotten math back to memory. ❤️🙏
@@leif1075 It comes from practicing algebra. I dont like the radical sign. So i always try to write it as an exponent instead. Example: sqrt(2)=2^1/2 or cube root of 2 as 2^1/3 or the 4th root as 2^1/4 ect.. Once you do that you will realize that you are multiplying exponents. In this case the exponents are fractions. And so you can multiply straight across. Example (((2)^1/2))1/3)1/4) And so the bottom of the fraction is just 2*3*4*5..... And since we didnt start at 0, we can subtract 2 from the factorial since we know that 0!=1 and 1!=1 Just make sure you look at the problem carefully. Calculus is all about algebra such that if you dont know your axioms and theorems, its going to be hard to solve quickly. Try seeing numbers as something besides their obvious value. Example: 4=2^2 or 120=5! Try understanding things on the most basic level. Thats what the greats do. From Aristotle to Feynman. Try to make it simple through understanding not memorization.
Im not sure why my reply didnt go through. RU-vid must have flagged for spam. I see it because i converted the radicals to exponents. Then realized that the bottom of the fraction (exponent) had a pattern. (1/2)(1/3)(1/4) Which is just 1/x! And we didnt start at 1 so we had to subtract 2 because... 0!=1 1!=1 Calculus is all about testing your understanding of algebraic axioms. Then, applying the theorems of calc. Hope this helps
ATTEMPT: The inside can be rewritten as x^1/2! * x^1/3! * x^1/4! * x^1/5 * ... which simplifies to x^(1/2! + 1/3! + 1/4! + 1/5! + ...) which again simplifies to x^(e-2). The integration is easy now! Int(x^(e-2) dx) = x^(e-1) * 1/(e-1) + C
did get a chuckle out of me when you wrote it down in a exponential series. I was wondering, "what does 1/2 + 1/2/3 + 1/2/3/4 + ..." converge too, and didn't think it's just factorial summation lol
I am extremely grateful that you uploaded this video for those who are educators and those who wish to learn mathematical concepts. You are amazing teacher. I wish I had teachers like you with great coherent voice and pace as well.
I didn't watch it past the first image of the problem, it looked pretty obvious we were getting the factorial reciprocals infinite sum missing n=0 and n=1, so you just had to substract 1/0! + 1/1! = 2 from e and then the integral is textbook indeed
Wow MIT question I like it. Although I've never solved any mit bee questions yet, just some qualifier ones😅. But Im still proud of myself since Im only in high school
At first I tried to do different ways of substitution, trying to arrive at a recursion or something nice, but it didn't play out well. Then I read what you wrote in the description of the video, and it suddenly became easy... 😃
Hello Mr. Can you make a video on all these important series one is expected to know? I figured out all the solution until I was stumped by the sum of the reciprocals of the factorials. Thanks 🙏
When you get to a series like that, especially with factorial denominators, it usually implies a Taylor (or MacLauren) expansion, which is generally of the form f(x)/0!+f'(x)/1!+f''(x)/2!+...+fⁿ(x)/n!+... To infinity, where fⁿ(x) is the n'th derivative of the function at point x. For particular "nice" functions, if you add up all infinite terms, you get a power series (infinite polynomial) that perfectly approximates the function itself in a region. Any time you see this kind of structure, it is useful to try and find the function f(x) which gives that particular expansion. In this case, e^x is its own derivative, so it is the "purest" expansion that doesn't require weird modification for each term, thus 1/0!+1/1!+1/2!...=1⁰/0!+1¹/1!+1²/2!+... Which is e¹ (e^x evaluated at 1)
I just did this one in my head! I can't remember the last time I've been able to do that with one of these :D x^ (1/2 + 1/6 + 1/24 + 1/120 + ...) = x^(e-1) ∫ x^(e-2) dx = x^(e-1) / (e-1) + C
It’s appalling, that after more than half a decade of having computers around us, there is still no good and standardized way to express and render mathematical formulae in native written text on the internet.
@@oniondeluxe9942yeah they really need to somehow add LaTeX formatting into RU-vid comments, maths is just another language in the end. All the math communities would benefit from it
@@adw1z perhaps. But LaTeX is a very programmer oriented interface. They should have been able to come up with something smoother by now. But, where there is no potential for advertising - no interest.
