The problem looks easy but the answer is unexpectedly fancy. Love this, it reminds me of good old days in my engineering college. Wish to see an application in real life soon.
And here I was thinking this was going to detour through power series. I did not recognize the pattern. What I did for the expression in the thumbnail was find a simpler expression for the sum inside the limit, which gave the alternating sum 1 - 1/2 + 1/3 - 1/4 + ... I happened to recall the value for this, but to prove it, I converted it to a power series, took the derivative (which gave the geometric series with common ratio - x), converted the derivative of the power series to 1/(1+x), integrated, established that the constant of integration was zero, and evaluated the antiderivative at x = 1.
For the second limit, we can also mutliply top and bottom by sqrt(x)+1 so the denominator becomes x-1. After a little bit of algebraic manipulations, we eventually get the limit definition for f(x)=x*sqrt(x+1) at x=1 and for g(x)=sqrt(x+1) at x=1 which results in the same value
Another nice way to evaluate the limit would be to notice that the sum equals (1+1/2+1/3+...+1/(2n))-(1+1/2+1/3+...+1/n). Then using the fact that the partial sums of the harmonic series grow like log x, we get log(2n)-log n=log 2.
In Analysis we were given the following problem lim(1/n*(1+sec^2(pi/4n)+sec^2(2*pi/4n)+…+sec^2(n*pi/4n))) as n goes to infinity For the longest time I couldn’t figure it out until I remembered the limit definition of the integral, then it just becomes an integral of sec^2, which is relatively straightforward.
First sum can also neatly be solved with the well-known inequalities 1-1/x ≤ ln(x) ≤ x - 1. If 1-1/x = 1/(n+k), then x = (n+k)/(n+k-1), therefore 1/(n+k) ≤ ln((n+k)/(n+k-1)). If x - 1 = 1/(n+k), then x = (n+k+1)/(n+k), therefore ln((n+k+1)/(n+k)) ≤ 1/(n+k). That means our sum can be squeezed between two telescoping logarithmic expressions with the same limits, works out quite neatly.
This was the form of a few homework problems in Stewart's Calculus text. He liked to see if you could recognize definite integral from the limit of the sums definition. Same for working backwards from the limit definition of the derivative. They can be tough if expression is "disguised algebraically"
Nice. In the limit example you can also multiply it with (sqrt(x^2+x)+sqrt2)/(sqrt(x^2+x)+sqrt2) . Then you get (x^2+x-2)/(sqrtx-1)(sqrt(x^2+x)+sqrt2) which is (x+2)(sqrtx+1)/(sqrt(x^2+x)+sqrt2) so the limit is 3/sqrt2
Just for registration, with respect to the second limit, you can multiply the numerator and denominator by (sqrt(x^2+x)+sqrt(2)). After some simplifications, we arrive at (x+2)*(sqrt(x)+1)/(sqrt(x^2+x)+sqrt(2)), whose limit is 6/(2*sqrt(2) )) =3/sqrt(2) as expected.
On the second on It is not necessary to get rid of the root at the denominator if you take u=√x and f(u)√(x⁴+x²), you can just differentiate that and it gets the same
In the second one, we can evaluate more directly by multiplying the numerator and denominator by the conjugate of each difference. This "frees up" the factor of x-1 in the numerator and denominator to be explicitly canceled, after which the limit can be evaluated by direct substitution
Yeah I don’t know why he ‘invented’ an (x-1) factor and had to do a strange (for a Calc 1 student) factoring when the conjugate was there for the taking.
Hey Michael, In our calculus one (MIPT, Russia) the teachers loved the problems with finding limits using high order taylor series (that is, 2 and higher). You might be interested in doing a video on the one like those; in fact, I'll try to find and drop you some by email just in case.
The limit in the second part of the video can be calculated, much faster and easier, by multiplying and dividing both by [sqr (x ^ 2 + x) + sqr (2)] and by [sqr (x) +1]. But the method you have shown is certainly interesting for similar limits involving transcendent functions.
I always forget the Riemann sums ! Another way of seeing this is because the partial sums of the harmonic sums 𝐻_n are equivalent to ln(𝑛) as n goes to infinity. The sum we are looking at can also be written H_(2n) - H_n and we have : H_(2n) - H_n = ln(2n) - ln(n) + O(1/n) = ln(2) + O(1/n) when n goes to infinity, which means that the sum converges to ln(2) as n goes to infinity.
I set this discrete sum equal to the continuous sum (from 1 to n) and set them equal to each other. My reasoning is somewhat unrigorous but since the gradient of 1/x approaches 0 as x grows without bound, the two sums are equal at infinity. From here it is easy to solve for the discrete sum.
The result R must be bounded between the integrals: n R ≥ ∫ 1/(n+x) = ln(2n)-ln (n+1) = ln(2n/(n+1)) 1 n-1 R ≤ ∫ 1/(n+x) = ln(2n-1)-ln (n) = ln((2n-1)/n) 0 When n approaches ∞, both values approach ln(2)
I saw this one the other day and just thought "oh digamma function" ψ(2n + 1) - ψ(n + 1) And then just remembered it's basically natural log as it goes to infinity. Screw all that work. log(2) and done
Gah I always forget that method. I squeezed the sum 1/(n+1) + 1/(n+2) +... + 1/(2n) between the integral of 1/x from n to 2n and the integral of 1/x from n+1 to 2n+1.
Hello Sir, What tool are you using for making thumbnail. Your video thumbnails are always awesome. Kindly tell me as I'm in struggling position on UA-cam
In the second problem couldn't we change the variable , setting u=sqrt(x)?The result would be a derivative on its own without the constant part outside.
The first sum is exactly the probability that a permutation of 2n elements has a cycle of length n+1 or longer. This calculation shows up in the (relatively) famous problem of the 100 prisoners and 100 boxes
my professor give me the reverse of this on Calc1 with answer but no solution and I spent 2 days finding a solution the question is “find integral x=1 to 5 1/x dx by using infinite sum”
The problem looks easy but the answer is unexpectedly fancy. Love this, it reminds me of good old days in my engineering college. Wish to see an application in real life soon.
Yes, these conceptual understanding problems are great. Love them!
This is very enlightening and nostalgic; it brought me back to my Calc 1 days in college.
Thank you, professor!
I am a math teacher and yes, I love these problems! Thanks for sharing!
And here I was thinking this was going to detour through power series. I did not recognize the pattern. What I did for the expression in the thumbnail was find a simpler expression for the sum inside the limit, which gave the alternating sum 1 - 1/2 + 1/3 - 1/4 + ... I happened to recall the value for this, but to prove it, I converted it to a power series, took the derivative (which gave the geometric series with common ratio - x), converted the derivative of the power series to 1/(1+x), integrated, established that the constant of integration was zero, and evaluated the antiderivative at x = 1.
For the second limit, we can also mutliply top and bottom by sqrt(x)+1 so the denominator becomes x-1.
After a little bit of algebraic manipulations, we eventually get the limit definition for f(x)=x*sqrt(x+1) at x=1 and for g(x)=sqrt(x+1) at x=1 which results in the same value
yes, that's what I thought he was going to do. Michael's way was less intuitive to me, but in the end it's all the same.
Another nice way to evaluate the limit would be to notice that the sum equals
(1+1/2+1/3+...+1/(2n))-(1+1/2+1/3+...+1/n).
Then using the fact that the partial sums of the harmonic series grow like log x, we get
log(2n)-log n=log 2.
I solved it the exact same way!
Nice
In Analysis we were given the following problem
lim(1/n*(1+sec^2(pi/4n)+sec^2(2*pi/4n)+…+sec^2(n*pi/4n))) as n goes to infinity
For the longest time I couldn’t figure it out until I remembered the limit definition of the integral, then it just becomes an integral of sec^2, which is relatively straightforward.
You're amazing, Michael. Very impressive reasoning and I admire your deft understanding of the subject matter.
9:00 Wishing you the ability to understand fast and easy, and the power to excel in your studies. Have a great week at school!
Hi,
4:47 : amazing!
For fun:
9:01 : nice photo composition with the splash screens.
First sum can also neatly be solved with the well-known inequalities 1-1/x ≤ ln(x) ≤ x - 1.
If 1-1/x = 1/(n+k), then x = (n+k)/(n+k-1), therefore 1/(n+k) ≤ ln((n+k)/(n+k-1)).
If x - 1 = 1/(n+k), then x = (n+k+1)/(n+k), therefore ln((n+k+1)/(n+k)) ≤ 1/(n+k).
That means our sum can be squeezed between two telescoping logarithmic expressions with the same limits, works out quite neatly.
This was the form of a few homework problems in Stewart's Calculus text. He liked to see if you could recognize definite integral from the limit of the sums definition. Same for working backwards from the limit definition of the derivative. They can be tough if expression is "disguised algebraically"
I miss the tapping of the blackboard and the filling in of the box before the next segment 'appears'!
yes, although as far as I know there isn't much more to them than recognizing the format.
Nice. In the limit example you can also multiply it with (sqrt(x^2+x)+sqrt2)/(sqrt(x^2+x)+sqrt2) . Then you get (x^2+x-2)/(sqrtx-1)(sqrt(x^2+x)+sqrt2) which is (x+2)(sqrtx+1)/(sqrt(x^2+x)+sqrt2) so the limit is 3/sqrt2
Just for registration, with respect to the second limit, you can multiply the numerator and denominator by (sqrt(x^2+x)+sqrt(2)). After some simplifications, we arrive at (x+2)*(sqrt(x)+1)/(sqrt(x^2+x)+sqrt(2)), whose limit is 6/(2*sqrt(2) )) =3/sqrt(2) as expected.
On the second on It is not necessary to get rid of the root at the denominator if you take u=√x and f(u)√(x⁴+x²), you can just differentiate that and it gets the same
All mathematical problems can be solved by applying the definition(s). A lemma is just a cookie cutter for shortcuts and abbreviations.
In the second one, we can evaluate more directly by multiplying the numerator and denominator by the conjugate of each difference. This "frees up" the factor of x-1 in the numerator and denominator to be explicitly canceled, after which the limit can be evaluated by direct substitution
Yeah, that's what my calculus teacher would've always told us to do in situations like this
Yeah I don’t know why he ‘invented’ an (x-1) factor and had to do a strange (for a Calc 1 student) factoring when the conjugate was there for the taking.
Hey Michael,
In our calculus one (MIPT, Russia) the teachers loved the problems with finding limits using high order taylor series (that is, 2 and higher). You might be interested in doing a video on the one like those; in fact, I'll try to find and drop you some by email just in case.
Yes. Like the one from the Arnold's mathematical trivium:
lim x-> 0 (sin tan x - tan sin x)/(arcsin arctan x - arctan arcsin x)
@@jakubkocak887 Arnold rules, and this particular one looks cool too!
The limit in the second part of the video can be calculated, much faster and easier, by multiplying and dividing both by [sqr (x ^ 2 + x) + sqr (2)] and by [sqr (x) +1]. But the method you have shown is certainly interesting for similar limits involving transcendent functions.
I always forget the Riemann sums !
Another way of seeing this is because the partial sums of the harmonic sums 𝐻_n are equivalent to ln(𝑛) as n goes to infinity.
The sum we are looking at can also be written H_(2n) - H_n and we have : H_(2n) - H_n = ln(2n) - ln(n) + O(1/n) = ln(2) + O(1/n)
when n goes to infinity, which means that the sum converges to ln(2) as n goes to infinity.
I set this discrete sum equal to the continuous sum (from 1 to n) and set them equal to each other. My reasoning is somewhat unrigorous but since the gradient of 1/x approaches 0 as x grows without bound, the two sums are equal at infinity. From here it is easy to solve for the discrete sum.
The result R must be bounded between the integrals:
n
R ≥ ∫ 1/(n+x) = ln(2n)-ln (n+1) = ln(2n/(n+1))
1
n-1
R ≤ ∫ 1/(n+x) = ln(2n-1)-ln (n) = ln((2n-1)/n)
0
When n approaches ∞, both values approach ln(2)
I saw this one the other day and just thought "oh digamma function"
ψ(2n + 1) - ψ(n + 1)
And then just remembered it's basically natural log as it goes to infinity. Screw all that work. log(2) and done
practiced this one !
Gah I always forget that method. I squeezed the sum 1/(n+1) + 1/(n+2) +... + 1/(2n) between the integral of 1/x from n to 2n and the integral of 1/x from n+1 to 2n+1.
Can we do first problem without integral?
for part 2 i factored the first term as sqrt(x)*sqrt(x+1) and then i set sqrt(x) to x and saw I could set f(x) = x*sqrt(x^2 + 1)....
Hello Sir,
What tool are you using for making thumbnail.
Your video thumbnails are always awesome.
Kindly tell me as I'm in struggling position on UA-cam
Can this be done without Riemann-sums?
In the second problem couldn't we change the variable , setting u=sqrt(x)?The result would be a derivative on its own without the constant part outside.
In South Korea, when I high school student, I have saw before this problem some math problem book as high difficultly problem
How did I submit Questions to You ?
For the limit, you could have simply set y=\sqrt{x}, and then the limit would be an obvious derivative of \sqrt(y^4+y^2).
Not really in surreal numbers
I think these are useful for review, even if they're simple once you get it
put a suit on, ted.
Nice
The first sum is exactly the probability that a permutation of 2n elements has a cycle of length n+1 or longer. This calculation shows up in the (relatively) famous problem of the 100 prisoners and 100 boxes
I hate infinite series. They aren’t so straightforward as integrals 😡🥺
Woww i solved it
Ln(2) 😃
i think most of the people aren't familiar with converting rienman sum into normal integral i always like to see these type of question
9:00
my professor give me the reverse of this on Calc1 with answer but no solution and I spent 2 days finding a solution
the question is “find integral x=1 to 5 1/x dx by using infinite sum”
Eh... child's play.
Here is a hard problem:
What is the nature of this sequence: sum from k=0 to k=n of 1/C(n,k)
Calculate it if possible 😄