Yeah, I got lost there trying to figure out how the theorem applied also on the integral bounds and then waited to see 0 to 1 twice in a row thereafter.
some quick intuition: tan(x) gives you the slope of a line that makes an angle x with the positive x axis. Thus, arctan(x) gives you the angle a line with slope x makes with the positive x axis (in (-pi/2 to pi/2) of course) If you reflect a line passing around y = x, the slope of that line becomes it's reciprocal. Putting this all together, take a line with angle t. The new line formed by reflecting it about y = x makes an angle of pi/2 - t with the positive x axis. Thus, the sum of their angles with the positive x axis is pi/2 This corresponds to arctan(x) + arctan(1/x) = pi/2
The derivative d/dx arctan(x) = 1/(1+x^2), but the derivative d/dx arccot(x) = -1/(1+x^2). Thus the derivative of their sum is d/dx [ arctan(x) + arccot(x) ] = 0. The sum itself then must be a constant. Evaluate at any x, i.e., use arctan(1)=arccot(1)=pi/4. Thus we show that arctan(x) + arccot(x) = pi/2. Now note that arccot(x) = y means x = cot(y);then 1/x = tan(y) and finally arctan(1/x) = y. Thus we show arccot(x) = arctan(1/x). Putting the two results together, we have arctan(x) + arctan(1/x) = pi/2.
I still think that the trick of writing a sum as an evaluated integral and then using the dominated convergence theorem to swap the sum and the integral around is so slick. I had never seen this method before and now I absolutely love it.
Reminds me of one of my favorite theorems from analysis. A convergent, non-absolutely convergent series can be rearranged to have any real limit. It's actually really easy to prove.
Your approach (using integrals) is impressive and highly effective! I appreciate how you showcase alternative methods to solve the problem. Another way is utilizing Taylor expansion for the arctan function. It's fascinating to see the diverse range of approaches available. Thanks for sharing your expertise and expanding our problem-solving toolkit!
for the arctan's in the end you can also consider that sqrt(2)+1=1/(sqrt(2)-1) and therefore arctan(sqrt(2)-1)=arccot(sqrt(2)+1) and hence we can just use the standard trigonometric identity arctan(x)+arccot(x)=pi/2 (just consider a right triangle with one angle arctan(x))
Little note: To avoid pulliing the x^4+1 factorization out of a hat, first write it as a product of four factors given from the four roots of -1, then look at the pairs of conjugate factors. These give the desired factorization into quadratics. You can almost do this in your head (at least after you have already done it on paper :). This method generalizes to find other kinds of real factorizations, starting with a complex factorization.
That's a nice way. Also, you can see it as a difference of squares factorization by writing x^4 + 1 = (x^4 + 2x^2 + 1) - 2x^2 = (x^2 + 1)^2 - (sqrt(2)x)^2
There's an easy way to evaluate that integral x²+1/(x⁴+1) Divide by x² (1+1/x²)/(1/x²+x²) (1+1/x²)/[(x-1/x)²+2] let u=x-1/x and the integral will become like this 1/(u²+2) 😇
Another way to solve it : For a complex z such that z^4=-1, compute the sum S(z) = sum z^n/n (for n =1 to +inf). We know that S(z)= integral(sum z^n, n=0 to +inf)= -log(1-z). Then, we split S(z) =S0(z)+S1(z)+S2(z)+S3(z) where Sk(z)= sum z^{4n+k}/(4n+k). We have S0(z)=S(z^4)/4=-log(2)/4 S2(z)=(1/2)*sum (z²)^{2n+1}/(2n+1) (for n=0 to +inf)= (1/2)(S(z²)-0.5*S(z^4))=-ln(1-z²)/4+log(2)/2 Let M(z)=S1(z)+S3(z)= sum (-1)^n (z/(4n+1)+z^3/(4n+3)) Let a=exp(i*pi/4) so a^2=i and 1-a=2sin(pi/8)exp(-i*pi/8) . By evaluating S(z) at a, we get: -log(1-a)=log(2)/4 -log(1-i)/4+M(a) where log(1-a)=log(2)+log(sin(pi/8))-i*3pi/8 and log(1-i) =log(2)/2-i*pi/4 . By taking the imaginary part, we get: sqrt{2}/2 * sum (-1)^n * (1/(4n+1)+1/(4n+3)) = pi/4, which concludes the proof.
The completion of the square comes about naturally from the process of getting the real factorization of x^4+1: The fourth roots of −1 are (1±i)/√2 and (−1±i)/√2, and the factors corresponding to the first two roots are x−1/√2−i/√2 and x−1/√2+i/√2, and their product is (x+1/√2)^2+½; the same result holds for the other two fourth roots of −1.
Newton was doing elliptical integrals 200 years before everyone else caught up. The reason why we stopped naming everything after Newton is because way too much stuff would be named after him. Taylor series.... Newton came up with those
Indians were using Taylor series before Newton, Mathologer just did a recent video on that called "Powell’s Pi Paradox: the genius 14th century Indian solution"
I got this sum when trying to solve the integral of1/(1+x^4) from 0 to infinity without "special" functions. Edit: funny you did the exact opposite. I wrote the first part before starting the video.
Set f(x)=(1+x^2)/(1+x^4) so I=Int(f(x),x=0..1) as in the video. Substituting x=1/t gives the same integrand with integral over the range 1 to infinity. Thus 2I=Int(f(x),x=0..infinity or 4I=Int(f(x),x=-infinity..infinity). Now we can close the contour in the upper half plane, since the integrand is sufficiently decaying and pick up the residues at x=exp(i*Pi/4) and x=-exp(-i*Pi/4). Then I=-i*Pi/8*(2i*sin(Pi/4) + 2i*sin(3Pi/4))=Pi/4*(1/sqrt(2)+1/sqrt(2))=Pi/(2*sqrt(2)).
I bet this thing converges super quickly. Speaking of convergence, aren't you playing a little fast and loose at 5:27, when you convert the sum of -x^4 to 1+x^4 when in the problem x actually equals 1? Or am i missing something?
The sum seems similar to the Leibniz formula for pi. Can those two sums be put together to get a series that converges slightly faster to a multiple of pi than the sums individually would?
Your video inspired me to look closer at the general sums of 1/2, 1/3, 1/4, cyclically multiplied with some signs/coefficients. Let a_0, a_1, ... a_{m-1} to be numbers, which the series [0, 1/1, 1/2, 1/3, 1/4, ...] is cyclically multiplied by to create the sum. For example, let a_0 = -1, a_1 = 1, and the resulting sum is -1·0 + 1/1 - 1/2 + 1/3 - 1/4 + ... This TH-cam video suggests `a` to be [0, 1, 0, 1, 0, -1, 0, -1] Sum of a_k is required to be zero, to make the sum converge. Let F_k to be the discrete Fourier tranform of a_{i}, F_k = Σ_{k=0}^{m-1} e^(-2π·k·l/m) · a_{l}, k∈[0, m-1]. So, F is sum of sin and cos of angles, which are multiple of 2π/m. The result sum value is: (1/m)·Σ_{k=1}^{m-1} ( F[k] · (-ln(2 - 2cos(2πk/m))/2 + (iπ/2) · (1 - 2k/m)) ) -ln(2 - 2cos(2πk/m))/2 + (iπ/2) · (1 - 2k/m) is explicit formula for ln(1-wᵏx) for complex logarithm. ln(1-wᵏx) is sum of infinite power series Σ xⁿ/n with x replaced to wᵏx. Where w is m-th root of unity. The whole formula derived from sums of ln(1-wᵏx) where k is changed from 0 to m-1, written as power series: the wᵏⁿxⁿ sums to zero everywhere, expect n divisible by m. If sum not just ln(1-wᵏx) but wᵏˡln(1-wᵏx) the non-zero coefficients in the sum over k=0..m-1 are shifted by l. Given that the result of this Fourier transform is conjugate-symmetric because input is real ( F_{m-k} = conjugate(F_k) ), this sum of complex numbers is a real value with zero imaginary part. It can be written explicitly as a real-valued sum if this sum is replaced by summing conjugate-symmetric summand pairs k and (m-k) The first summand with zero index is missing, because sum of a_k is required to be zero to make the sum convergent, and this zero sum is F[0]. This sum result for integer `a` can be described like this: Let R to be numbers, composed from linear combination sums of sin(2pi·k/m) and cos(2pi·k/m) with rational coefficients. Let Q to be the set of ln(2 - 2cos(2πk/m)) numbers. They don't intersect with ln(2) when m is not multiple of 4, otherwise ln(2) is just one of them and two groups are mixed. Result is the sum of: - number from R multiplied by π, - number from R multiplied by ln(2), - numbers from R multiplied by numbers from Q. The full python code to define this sum: from sympy import * a = [0, 1, 0, 1, 0, -1, 0, -1] m = len(a) assert sum(a) == 0, "sum of coefficients must be zero to make the series to converge" F = [sum(a[l] * exp(-2*pi*I*k*l/m) for l in range(m)) for k in range(m)] result = sum(F[k] * (-ln(2 - 2*cos(2*pi/m*k))/2 + I*pi/2*(1 - Integer(k)*2/m) ) for k in range(1, m)) / m print(simplify(result.expand(complex=True))) Actual code can be made more sophisticated, to simplify symmetrical summands ln(1-cos(x)) + ln(1+cos(x)) as 2·ln(sin(x)) for more complicated `a` coefficients. sympy can't simplify ln(sqrt(2) + 1)/2+ln(sqrt(2) - 1)/2 to zero in long complicated expressions. Also the code can be optimized to utilize the conjugate-symmetry of F.
I invented this way to transform existing power series to their version with changed signs or cyclically multiplied by some coefficients. But I expect it was invented hundreds years ago. I am curious to learn what is actual name for this method.
14:49 I don't think this is true. I can see how, if you're being really cautious, you might compute the integral by taking limits at the endpoints when you're uncertain of the value. And certainly, doing so here can help you stumble into that next step you took to compute that value. But here there was never anything iffy about that value, so the caution at the integral step is not relevant to the step that follows. No, you take the limit from below, because that's what you need in order to take the sum of arctans here. No property of the integral step is relevant here. You have a definite number, and to compute its value, you find a limit that equals it (because arctan is continuous at that point) and use the extra wiggle room to perform a useful manipulation. In a different case, if the manipulation you wish to perform had requirements more strict than just "approach the point from below", you could well end up taking a very specific sequential limit, and it wouldn't make a difference to the reasoning here. The general limit evaluates to the desired number, so you may take any limit and be confident that it will evaluate to the number.
This was very very cludgy - two transforms as given, then domain suppression - it works. Why it works is not clear :(, some problems with elementary number theory here.
@@AlexandrBorschchev yea that's the sad reality ,if math and physics channels had got 100m views or so then we could see whole another level of content in these areas
he filmed a lot of videos while he was a bit sick. I think there's still a couple more coming up wherein he'll sound like this. To be very clear, he's okay now. These are recorded far in advance. :) -Stephanie, MP Editor
I think you need a bit of context. Michael does rock climbing with a club in CO. Coincidently, I'm from CO myself and know that Rifle is the name of a town in CO (you can learn more about Rifle, CO here: www.rifleco.org) and one he's probably been to for rock climbing. that explains the rocks in the shapes of mountains accompanying the shirt. The only connections to "guns" is that the city of Rifle gets its name from a local story (myth), that a *trapper had once left his rifle along a creek in the region*; from then on, the city has been known as Rifle. Feel free to edit your comment since it wouldn't be likely that you'd know the above before making the comment but now that you do, I'll expect better from you. Thanks Stephanie MP Editor, Producer
There was an error at 4:36. The upper bound of the integral is 1 not ∞. At 6:53, the bound was restored to 1.
Good job 👍😊😊
Oh that makes sense, I was confused how would the geom. series converge.
Was wondering if the bound going to infinity was a mistake and it was corrected by another 'mistake' to restore it.
Yeah, I got lost there trying to figure out how the theorem applied also on the integral bounds and then waited to see 0 to 1 twice in a row thereafter.
I was so confused when he wrote that down thanks for letting us know
the last part is overcomplicating things, sqrt(2)+1 and sqrt(2)-1 are (positive) reciprocals of each other so their arctangents add to pi/2
Proof?
@@rayniac211 1/(sqrt(2)-1)=(1+sqrt(2))/(sqrt(2)^2-1^2)=1+sqrt(2)
some quick intuition:
tan(x) gives you the slope of a line that makes an angle x with the positive x axis.
Thus, arctan(x) gives you the angle a line with slope x makes with the positive x axis (in (-pi/2 to pi/2) of course)
If you reflect a line passing around y = x, the slope of that line becomes it's reciprocal.
Putting this all together, take a line with angle t. The new line formed by reflecting it about y = x makes an angle of pi/2 - t with the positive x axis.
Thus, the sum of their angles with the positive x axis is pi/2
This corresponds to arctan(x) + arctan(1/x) = pi/2
The derivative d/dx arctan(x) = 1/(1+x^2), but the derivative d/dx arccot(x) = -1/(1+x^2). Thus the derivative of their sum is d/dx [ arctan(x) + arccot(x) ] = 0. The sum itself then must be a constant. Evaluate at any x, i.e., use arctan(1)=arccot(1)=pi/4. Thus we show that arctan(x) + arccot(x) = pi/2. Now note that arccot(x) = y means x = cot(y);then 1/x = tan(y) and finally arctan(1/x) = y. Thus we show arccot(x) = arctan(1/x). Putting the two results together, we have arctan(x) + arctan(1/x) = pi/2.
@@rayniac211 In any right triangle, the tangent of one angle is the cotangent of its complement
I still think that the trick of writing a sum as an evaluated integral and then using the dominated convergence theorem to swap the sum and the integral around is so slick. I had never seen this method before and now I absolutely love it.
Kinda reminds me of the feynman method.
But you have to verify that you got the conditions to apply the dominated convergence theorem.
By the way, the statement of the dominated convergence theorem does not include infinite series as those that are being treated here.
The GOAT is back.
Reminds me of one of my favorite theorems from analysis. A convergent, non-absolutely convergent series can be rearranged to have any real limit. It's actually really easy to prove.
Your approach (using integrals) is impressive and highly effective! I appreciate how you showcase alternative methods to solve the problem. Another way is utilizing Taylor expansion for the arctan function. It's fascinating to see the diverse range of approaches available. Thanks for sharing your expertise and expanding our problem-solving toolkit!
for the arctan's in the end you can also consider that sqrt(2)+1=1/(sqrt(2)-1) and therefore arctan(sqrt(2)-1)=arccot(sqrt(2)+1) and hence we can just use the standard trigonometric identity arctan(x)+arccot(x)=pi/2 (just consider a right triangle with one angle arctan(x))
Little note: To avoid pulliing the x^4+1 factorization out of a hat, first write it as a product of four factors given from the four roots of -1, then look at the pairs of conjugate factors. These give the desired factorization into quadratics. You can almost do this in your head (at least after you have already done it on paper :). This method generalizes to find other kinds of real factorizations, starting with a complex factorization.
That's a nice way. Also, you can see it as a difference of squares factorization by writing x^4 + 1 = (x^4 + 2x^2 + 1) - 2x^2 = (x^2 + 1)^2 - (sqrt(2)x)^2
There's an easy way to evaluate that integral
x²+1/(x⁴+1)
Divide by x²
(1+1/x²)/(1/x²+x²)
(1+1/x²)/[(x-1/x)²+2]
let u=x-1/x and the integral will become like this
1/(u²+2) 😇
You crushed it!
Another way to solve it : For a complex z such that z^4=-1, compute the sum S(z) = sum z^n/n (for n =1 to +inf). We know that S(z)= integral(sum z^n, n=0 to +inf)= -log(1-z).
Then, we split S(z) =S0(z)+S1(z)+S2(z)+S3(z) where Sk(z)= sum z^{4n+k}/(4n+k).
We have S0(z)=S(z^4)/4=-log(2)/4
S2(z)=(1/2)*sum (z²)^{2n+1}/(2n+1) (for n=0 to +inf)= (1/2)(S(z²)-0.5*S(z^4))=-ln(1-z²)/4+log(2)/2
Let M(z)=S1(z)+S3(z)= sum (-1)^n (z/(4n+1)+z^3/(4n+3))
Let a=exp(i*pi/4) so a^2=i and 1-a=2sin(pi/8)exp(-i*pi/8) . By evaluating S(z) at a, we get:
-log(1-a)=log(2)/4 -log(1-i)/4+M(a)
where log(1-a)=log(2)+log(sin(pi/8))-i*3pi/8 and log(1-i) =log(2)/2-i*pi/4 .
By taking the imaginary part, we get: sqrt{2}/2 * sum (-1)^n * (1/(4n+1)+1/(4n+3)) = pi/4, which concludes the proof.
love the new style and animations, so vibrant
Integral(0,1)=integral(1,∞)
→integral(0,1)=integral(0,∞)/2
Integral(0,∞)...by residue theorem, x^4=t and use ramanujan's master theorem, etc...
Me: *slips out calculator to find two arctan values
Penn: don't you dare cross the LIMIT.
The completion of the square comes about naturally from the process of getting the real factorization of x^4+1: The fourth roots of −1 are (1±i)/√2 and (−1±i)/√2, and the factors corresponding to the first two roots are x−1/√2−i/√2 and x−1/√2+i/√2, and their product is (x+1/√2)^2+½; the same result holds for the other two fourth roots of −1.
Wow, what a great resolution of this problem using math tricks and MP's calculating cunnig👍👏
Newton was doing elliptical integrals 200 years before everyone else caught up. The reason why we stopped naming everything after Newton is because way too much stuff would be named after him. Taylor series.... Newton came up with those
Indians were using Taylor series before Newton, Mathologer just did a recent video on that called "Powell’s Pi Paradox: the genius 14th century Indian solution"
The same with Euler. So many theorems named after him.
@@ingiford175 fake news
Chalk, board, and viewers that don't complain about trivial errors. That is how it used to be, that's how it should be.
Congrats on that little helper info!
Please note: At the 4.39 mark. the upper limit of the integral is 1 and not infinity.
I got this sum when trying to solve the integral of1/(1+x^4) from 0 to infinity without "special" functions.
Edit: funny you did the exact opposite. I wrote the first part before starting the video.
5:48 Won't there be a problem with interval of convergence ?
Maybe not but there is mistype
I can not hear you what is the problem
In the last minutes, you can consider Arctan(sqrt(2)+1)+Arctan(sqrt(2)-1)=pi/2
A bit overenthusiastic on the infinities on the bounds of integration on the first board? (the integrals of the sums should be from 0 to 1.)
At 14:58 I think you meant to say "it's an upper bound of integration"...
Is it possible to solve it just using taylor series or something easier
It doesn't involve the alternating harmonic sum, does it?
When you get to ∫ (i+x^2)/(1+x^4) dx, I really wanted to see some complex analysis.
At the very least, the complex factorization of x^4+1 does provide a major hint toward the real factorization.
Set f(x)=(1+x^2)/(1+x^4) so I=Int(f(x),x=0..1) as in the video. Substituting x=1/t gives the same integrand with integral over the range 1 to infinity. Thus 2I=Int(f(x),x=0..infinity or 4I=Int(f(x),x=-infinity..infinity). Now we can close the contour in the upper half plane, since the integrand is sufficiently decaying and pick up the residues at x=exp(i*Pi/4) and x=-exp(-i*Pi/4). Then
I=-i*Pi/8*(2i*sin(Pi/4) + 2i*sin(3Pi/4))=Pi/4*(1/sqrt(2)+1/sqrt(2))=Pi/(2*sqrt(2)).
Vote up, for good idea, and for using Maple sytax. 😀
I bet this thing converges super quickly. Speaking of convergence, aren't you playing a little fast and loose at 5:27, when you convert the sum of -x^4 to 1+x^4 when in the problem x actually equals 1? Or am i missing something?
I think you are missing the 1/1-r convergence of an infinite geometric sum. Here r is -x^4, so 1/1+x^4.
THANK YOU FROR PERSISTENCE EACH DETAIL
The sum seems similar to the Leibniz formula for pi. Can those two sums be put together to get a series that converges slightly faster to a multiple of pi than the sums individually would?
Your video inspired me to look closer at the general sums of 1/2, 1/3, 1/4, cyclically multiplied with some signs/coefficients.
Let a_0, a_1, ... a_{m-1} to be numbers, which the series [0, 1/1, 1/2, 1/3, 1/4, ...] is cyclically multiplied by to create the sum.
For example, let a_0 = -1, a_1 = 1, and the resulting sum is -1·0 + 1/1 - 1/2 + 1/3 - 1/4 + ...
This TH-cam video suggests `a` to be [0, 1, 0, 1, 0, -1, 0, -1]
Sum of a_k is required to be zero, to make the sum converge.
Let F_k to be the discrete Fourier tranform of a_{i}, F_k = Σ_{k=0}^{m-1} e^(-2π·k·l/m) · a_{l}, k∈[0, m-1]. So, F is sum of sin and cos of angles, which are multiple of 2π/m.
The result sum value is:
(1/m)·Σ_{k=1}^{m-1} ( F[k] · (-ln(2 - 2cos(2πk/m))/2 + (iπ/2) · (1 - 2k/m)) )
-ln(2 - 2cos(2πk/m))/2 + (iπ/2) · (1 - 2k/m) is explicit formula for ln(1-wᵏx) for complex logarithm. ln(1-wᵏx) is sum of infinite power series Σ xⁿ/n with x replaced to wᵏx. Where w is m-th root of unity. The whole formula derived from sums of ln(1-wᵏx) where k is changed from 0 to m-1, written as power series: the wᵏⁿxⁿ sums to zero everywhere, expect n divisible by m. If sum not just ln(1-wᵏx) but wᵏˡln(1-wᵏx) the non-zero coefficients in the sum over k=0..m-1 are shifted by l.
Given that the result of this Fourier transform is conjugate-symmetric because input is real ( F_{m-k} = conjugate(F_k) ), this sum of complex numbers is a real value with zero imaginary part. It can be written explicitly as a real-valued sum if this sum is replaced by summing conjugate-symmetric summand pairs k and (m-k)
The first summand with zero index is missing, because sum of a_k is required to be zero to make the sum convergent, and this zero sum is F[0].
This sum result for integer `a` can be described like this:
Let R to be numbers, composed from linear combination sums of sin(2pi·k/m) and cos(2pi·k/m) with rational coefficients.
Let Q to be the set of ln(2 - 2cos(2πk/m)) numbers. They don't intersect with ln(2) when m is not multiple of 4, otherwise ln(2) is just one of them and two groups are mixed.
Result is the sum of:
- number from R multiplied by π,
- number from R multiplied by ln(2),
- numbers from R multiplied by numbers from Q.
The full python code to define this sum:
from sympy import *
a = [0, 1, 0, 1, 0, -1, 0, -1]
m = len(a)
assert sum(a) == 0, "sum of coefficients must be zero to make the series to converge"
F = [sum(a[l] * exp(-2*pi*I*k*l/m) for l in range(m)) for k in range(m)]
result = sum(F[k] * (-ln(2 - 2*cos(2*pi/m*k))/2 + I*pi/2*(1 - Integer(k)*2/m) ) for k in range(1, m)) / m
print(simplify(result.expand(complex=True)))
Actual code can be made more sophisticated, to simplify symmetrical summands ln(1-cos(x)) + ln(1+cos(x)) as 2·ln(sin(x)) for more complicated `a` coefficients. sympy can't simplify ln(sqrt(2) + 1)/2+ln(sqrt(2) - 1)/2 to zero in long complicated expressions.
Also the code can be optimized to utilize the conjugate-symmetry of F.
I invented this way to transform existing power series to their version with changed signs or cyclically multiplied by some coefficients. But I expect it was invented hundreds years ago.
I am curious to learn what is actual name for this method.
I would have expected the residue theorem.
How newton came up with this result? 🤔
Being Newton, he thought very hard, and then wrote down the answer.
Great video and all, but isn't this series conditionally convergent?
Shocking news Isaac Newton has reincarnated and is now Mike Penn!
Wow
Boss!
What if you considered the sum
as 4/5 + 4/21 + ..., so that all
the terms are positive?
This kind of sum can come in JEE,even i have solved some of these
14:49 I don't think this is true.
I can see how, if you're being really cautious, you might compute the integral by taking limits at the endpoints when you're uncertain of the value. And certainly, doing so here can help you stumble into that next step you took to compute that value. But here there was never anything iffy about that value, so the caution at the integral step is not relevant to the step that follows.
No, you take the limit from below, because that's what you need in order to take the sum of arctans here. No property of the integral step is relevant here. You have a definite number, and to compute its value, you find a limit that equals it (because arctan is continuous at that point) and use the extra wiggle room to perform a useful manipulation.
In a different case, if the manipulation you wish to perform had requirements more strict than just "approach the point from below", you could well end up taking a very specific sequential limit, and it wouldn't make a difference to the reasoning here. The general limit evaluates to the desired number, so you may take any limit and be confident that it will evaluate to the number.
sweet
Please make a video how summing rational numbers can be an irrational number.
If u divide the numerator and denominator by x^2 and substitute u=x-1/x u get the derivative of u in the numerator directly
One of the variables in your post is u. There is a word "you" that needs to be spelled out. This is not partial texting.
This was very very cludgy - two transforms as given, then domain suppression - it works.
Why it works is not clear :(, some problems with elementary number theory here.
I think higher than 1080p video upload for these videos are not necessary. u just offer people bandwidth consumption
see something mathematically remarkable? It's probably called Euler's or Newton's something
You lost me at the end. :-(
Hummm.... I'm not sure with the integration and rearranging how could we justify that this isn't just another example of Riemann rearrangement?
Math and physics channels should get more views in youtube........😤😤😤😤 but most of the views goes to useless stuffs
satisfying crushing candies with car gets 100M views
@@AlexandrBorschchev yea that's the sad reality ,if math and physics channels had got 100m views or so then we could see whole another level of content in these areas
You all right there mate? Your voice sounds a little hoarse in this one
he filmed a lot of videos while he was a bit sick. I think there's still a couple more coming up wherein he'll sound like this. To be very clear, he's okay now. These are recorded far in advance. :)
-Stephanie, MP Editor
Sir, with the greatest of respect, why did you choose to wear such an offensive shirt, or do you agree with the mass slaughter of minors?
I think you need a bit of context. Michael does rock climbing with a club in CO. Coincidently, I'm from CO myself and know that Rifle is the name of a town in CO (you can learn more about Rifle, CO here: www.rifleco.org) and one he's probably been to for rock climbing. that explains the rocks in the shapes of mountains accompanying the shirt. The only connections to "guns" is that the city of Rifle gets its name from a local story (myth), that a *trapper had once left his rifle along a creek in the region*; from then on, the city has been known as Rifle.
Feel free to edit your comment since it wouldn't be likely that you'd know the above before making the comment but now that you do, I'll expect better from you.
Thanks
Stephanie
MP Editor, Producer
Series quite frankly sucks