what does "semi-stable" elliptic curve mean? How does it differ or same from elliptic curve? As a photographer, I find it really interesting about Fermat's Last Theorem that something as abstract as number theory problem could have visual structure. Beautiful!
@@cepatwaras The property of being "semi-stable" applies to elliptic curves defined over number fields or non-archimedean local fields. It means roughly that after reducing your curve modulo p (so that you have a curve over a finite field), the resulting curve is non-singular or has at worst an ordinary double point. This is probably not a very helpful explanation but see Silverman's book _Arithmetic of Elliptic Curves_. Much more general versions of "stable" and "semi-stable" exist for curves, their models, and their reductions in the language of scheme theory (see Qing Liu's _Algebraic Geometry and Arithmetic Curves_). It's these ideas that are paramount in Wiles' proof where one considered the Tate module of the Jacobian of a curve which induces a Galois representation, and semi-stability plays a crucial role here.
@@agrajyadav2951 once I saw the solution I realised it was yes, it was quite elegant to be fair, but there was one step I could get my head around which meant I couldn’t access the other parts. For those interested, I got rejected from ICL, the university I did the MAT for but I have from Cambridge, so don’t worry if things don’t go your way, you can still do it haha! :D
My take on part 2 (using calculus): We can differintiate both sides as such: dy/dx (2y - 1) = 3x^2 - 1. Dividing 2y - 1 both sides, we have dy/dx = (3x^2 - 1)/(2y - 1). Since y value on the graph has an undefined slope (vertical), we just set the denominator of dy/dx equal to 0: 2y - 1 = 0. Thus, we have y = 1/2.
@@SpartaSpartan117 is the graph not necessarily vertical there? What else can happen when a derivative approaches infinity? Not being an ass just genuinely interest
At the two points (gamma, delta) or (alpha, delta), the slope of the tangent lines satisfy the property dx/dy=0, so I just implicit differentiate the equation to find dx/dy=(2y-1)/(3x^2-1), and set y=1/2.
@@michaelayeni177 only for maths majors, in the UK you apply for a specific course, this exam is used as an admissions test for those applying to: maths, maths and philosophy, maths and stats, computer science, maths and computer science, computer science and philosophy. There are 7 questions but the applicants do 5 depending on what course they are applying for
Enjoyed the Question and ofcourse the video too. You should also have a look at the JEE advanced paper. It has some nice Questions in calculus and other math in general, could be good inspiration for your channel's audience!
Hi love ur vids, I had a request. Could you plz make some videos on the STEP Exam. It is the Cambridge entrance exam for High School students and the questions in it are absolutely brutal. There are 3 papers in total and they increase in difficulty from STEP 1 being the easiest and STEP 3 is the hardest. I think you will really enjoy some of the questions since in my opinion, having looked at both the JEE advance maths questions and STEP 3, I would say that a lot of the STEP 3 are actually even harder than Jee so plz give it a go. But if you want the really hard ones then do STEP 3. STEP 2 AND 1 are still hard but not as hard. Also, if you want the hardest ones, even from STEP 3, then there is a mark scheme which has all the answers of the questions from the past years and towards the end of that, you can go to STEP 3 and they tell you which questions were done well or poorly. Hence which were hardest and which were easiest.
@@blackpenredpen Also a second note if you google “STEP support programme” and click the second link down you should see all the foundation modules they have many problems there.
@@blackpenredpen I'm planning on doing this on my channel soon - how did you read my mind?! 😂 PS, great video - I can't think of a better way of spending Valentine's Day!
When I did this a few months ago I wasn’t able to do parts v and vi, the others ones I got correct but the last 2 where hard. I could’ve known v when I saw how it was done but vi was too hard for me
I was not familiar with Vieta's formulas so the explanation for the last part of section vi left me very confused. the missing piece was (-a sub n-1 / a sub n) = -1 / 1 = the sum of roots, where a sub i is the coefficient of the ith term in the cubic equation
for part 3 you can say that if (a,b) is on the curve then (a, 1-b) is on the curve and the distance to the line y=1/2 from each is abs(b-1/2) without writing any equations. Once I wrote it it is probably not faster than what you did.
set y = 0. x intecpt values∈{ -1,0,1} . due to the restrictions, the values are 1,-1. set x = 0. y intercept values∈{ 0,1} y is a parabola, since parabola's vortex is midway between the y-intercepts, the vortex of the graph is symmetric on the line y=0.5 since parabolas are symmetric on two sides of the vortex and the other side is independent of y, the graph is symmetric on the line y=0.5. (there are some circular reasoning that I'm not energized enough to resolve) since the three numbers already exist on the same horizontal line y=0.5, replace y with 0.5 in the original equation and move everything to one side gives x^3-x+0.25=0. replace the 0 with a new y gives a cubic equation that has the correct roots. the equation of the circle can be written as y=sqrt( (b-a)^2/4+(x-(b+a)/2)^2 )+0.5 which can be simplified to sqrt( - (x-a) (x-b) )+0.5 sub it into the "y" of the original problem can probably yield the correct answer. however, the amount of calculation already feels like this is the wrong path. however, by continuing the next step, the roots actually cancel out nicely to 0 = x^{3}+x^{2}-(1+a+b)x+(ab+0.25). then I got stuck since I don't know the cubic roots formula. so maybe this is indeed the wrong path. try again and define the circle as a function of y(right half) might work, but i really need to be doing my homework by now.
try this: elliptic integral of the 2nd kind, integrate sqrt(1+c*sin^2(z) ] dz = (2/3)*csc^2(z)*(c*sin^2(z)+1)^(3/2), the c-value is negative, csc has a zero-point issue
This was easier than I thought it was.....I was able to solve most of of except the 6th on my own....although in the exam hall scenario I would panicked and definitely skip it.....it was an excellent problem tho
I once did a paper which purely consited of greek letters, and when I was doing it, I gave up because as I progressed, all of the letters just began to look identical
Hey blackpenredpen! Can u solve the following integral: ∫(x³ - 4x² + 3x + 6)/√(x² + 3x + 5) dx evaluated from 0 to 4. I challenge you to solve this without using any of the hyperbolic functions or their inverses... use the logarithm instead... By the way, great video as always!! I am a huge fan of yours!
I did the Oxford maths entrance exam in 1993, and scored 98% on the pure paper. The funny thing was that loads of other people got very similar marks. They’d just made the paper a little too easy / similar to previous years. In contrast very strong candidates were getting in with very low scores (like 40%) the year before - the 1992 paper was just a bit too hard. The margins can be quite small at this level for a bunch of pretty well prepared candidates.
@blackpenredpen i am thirsty, not for water, not for integrals, not for even the most delightfull of elliptic curves. I am thirsty for some beautiful authentic bprp socks. thankyou for your content, i just must open my wallet at long last to purchase some math for fun socks. If you could drop some merch socks i would be so thankful kind sir :))
There is an equation that popped into my mind and I still don't know how to solve it. Sin (x)=x+5 Of course, x needs to be negative because sin is only between -1 and 1 or maybe the answer is a complex number but I still don't know how to approach it. According to WolframAlpha, the answer is -4.15252 but I don't know how they got the answer I think they plotted both graphs and found the intersection point but how can we solve it algebraically. Please help
Hey I remember answering this question, it was a pretty nice question tbh! Btw i ended up geting 30 points out of 100 which is kinda crappy but hey i enjoyed the test and i appreciate the opportunity at least
At 2:56, I used implicit differentiation to solve for the y-value. I prefer to, and it is easier for me lol. Differentiate both sides with respect to y and also got the answer δ=0.5. (Although a bit more working out)
For gamma since the curve is symmetric across the value of gamma, I plug in b for x. Then I solve for both y's. The average of those y's is gamma. And you get gamma=1/2.
Can you guys solve this, this is from peterson related rates problem one ship is sailing south at a rate of 5 knots, and another ship is sailing east at a rate of 10 knots. at 2 P. M. the second ship was at the place occupied by the first ship one hour before. at what time was the distance between the ships not changing?
My take on part 2 with some calculus: Let g : (-1,1) -> R^2 be a one to-one differentiable path with the image of g completely within the given curve, furthermore suppose that g' is never zero, and suppose g(0)=(x_0,δ) where x_0 is one of α,β,γ. Note: since F(x,y)=x^3-x-y^2+y is continuously differentiable and ∇F(x,y)=(3x^2-1, -2y+1), we have that the conditions for the implicit function theorem (for extraction of one of the variables as a function of the second) hold everywhere except for the points where y=1/2, x=±√1/3, in particular those points are not on the curve, and so such a path does indeed exist for any point on the curve that we choose. Now after using a paragraph to justify the existence of g, we know by geometric arguments that at the point g(0) we have that g'(0)=(0,y'(0)) for some nonzero y'(0), but writing g(t)=(x(t),y(t)) and differentiating the curve's equation (which holds for the image of g): (2y(t)-1)y'(t)=(3x(t)^2-1)x'(t) and in particular (2y(0)-1)y'(0)=0 and since y'(0) is nonzero, we have y(0)=1/2, so δ=1/2
I'm just guessing at this point (only 2.38 into the video) but y=delta is a line of symmetry and on the lhs we have y^2-y=y*(y-1) so there is a symmetry mid way between y=0 and y=1. (because if you think about it (swapping y and y-1 would leave the equation unchanged)) (lol not only was i right but answered the next question as well.)
Isn't it much easier for proofing the symmetry to replace y with 1/2 - (y - 1/2)? Simplify to 1 - y. f(x, y) := y^2 - y = x^3 - x. Then f(x, 1 - y) is (1 - y)^2 - (1 - y) = x^2 - x => 1 - 2y + y^2 - 1 + y x^2 - x => y^2 - y = x^3 - x
I think I have a cleaner explanation of why the curve is symmetric about the line. For any point (X,Y) on the curve we know Y^2 - Y = X^2 - X. Now consider the point (X,Y') which is (X,Y)'s reflection about the line y = 1/2. We have (Y + Y')/2 = 1/2 since the midpoint must be on the line, so Y' = 1 - Y. Therefore Y'^2 - Y' = (1 - Y)^2 - (1- Y) = 1 - 2Y + Y^2 - 1 + Y = Y^2 - Y = X^2 - X. So the point (X,Y') also lies on the curve and hence the curve is symmetric about y = 1/2.
this is random but how would you find the area of the little circle/oval/ellipsoid thingy on the left? y cant be defined explicitly in terms of x in a nice way, and i dont think polar coordinates would work here
Because you are finding the distance from beta to alpha. To do this you minus alpha from beta. Then divide by two for radius. Like if you had 2 and -1. The distance would be 2-(-1)= 3 . Alpha + beta / 2 finds the arithmetic average between the two points with corresponds to a number on the x axis which is the distance from the y axis.
If the quadratic equation can be solved use the formula, namely the formula abc. can cubic equations also be solved using formulas ?, if you can use what formulas?
Why is my math soooo rigorous even in high school? Like soooo many epsilon delta proofs (we use alpha and beta instead of epsilon and delta xD) and many techniches for reasoning, not to mention the EXTREMELY mind bending geometry, our math curriculum is originally french, so maybe that's why.
He's using (ay + b)^2 = (a^2)(y^2) + 2aby + b^2 in reverse, picking values of a and b that will end up as (y^2 - y) plus some constant. So a^2 must be 1 and 2ab must be -1. Choose a = 1 and b = -1/2 to satisfy this, so b^2 is 1/4. Then observe that (y - 1/2)^2 = (y^2 - y + 1/4) according to the formula I gave in the beginning, so that's what you want to achieve on the right-hand side.
![The Most Useful Curve in Mathematics [Logarithms]](/img/n.gif)
Fermat's Last Theorem is proved with Elliptical Curves so I am excited.
Omg same!
what does "semi-stable" elliptic curve mean? How does it differ or same from elliptic curve?
As a photographer, I find it really interesting about Fermat's Last Theorem that something as abstract as number theory problem could have visual structure. Beautiful!
@@cepatwaras The property of being "semi-stable" applies to elliptic curves defined over number fields or non-archimedean local fields. It means roughly that after reducing your curve modulo p (so that you have a curve over a finite field), the resulting curve is non-singular or has at worst an ordinary double point. This is probably not a very helpful explanation but see Silverman's book _Arithmetic of Elliptic Curves_. Much more general versions of "stable" and "semi-stable" exist for curves, their models, and their reductions in the language of scheme theory (see Qing Liu's _Algebraic Geometry and Arithmetic Curves_). It's these ideas that are paramount in Wiles' proof where one considered the Tate module of the Jacobian of a curve which induces a Galois representation, and semi-stability plays a crucial role here.
@IonRubywho asked?
ELLIPTIC curves please.
This was the one question I couldn’t solve hahaha I saw the elliptic curve and thought oh no 🤣
Did you get an offer?
Explain what is this admission test ?
@@azeds It's the admissions test to study undergraduate mathematics at Oxford.
It was pretty damn easy ngl
@@agrajyadav2951 once I saw the solution I realised it was yes, it was quite elegant to be fair, but there was one step I could get my head around which meant I couldn’t access the other parts. For those interested, I got rejected from ICL, the university I did the MAT for but I have from Cambridge, so don’t worry if things don’t go your way, you can still do it haha! :D
I sat this exam last year for Oxford/Imperial and I absolutely loved this question! Very exciting seeing you cover it !! :)
My take on part 2 (using calculus): We can differintiate both sides as such: dy/dx (2y - 1) = 3x^2 - 1. Dividing 2y - 1 both sides, we have dy/dx = (3x^2 - 1)/(2y - 1). Since y value on the graph has an undefined slope (vertical), we just set the denominator of dy/dx equal to 0: 2y - 1 = 0. Thus, we have y = 1/2.
That’s a good one.
That was my first thought on how to solve it as well. Although we don't have a guarantee that the slope is really vertical there
@@SpartaSpartan117 Nice! Also thank you @blackpenredpen
Brilliant.
@@SpartaSpartan117 is the graph not necessarily vertical there? What else can happen when a derivative approaches infinity? Not being an ass just genuinely interest
At the two points (gamma, delta) or (alpha, delta), the slope of the tangent lines satisfy the property dx/dy=0, so I just implicit differentiate the equation to find dx/dy=(2y-1)/(3x^2-1), and set y=1/2.
I believe that Greek letter at 16:32 is a lower case squigma.
I did this question in the actual exam and it was definitely the one I had most doubts about
@zlnst6 I did:))
@@diffusegd hey what did u get in the MAT if you don’t mind me asking
Is this an exam for all oxford applicants or only for math majors interested in Oxford?
@@michaelayeni177 only for maths majors, in the UK you apply for a specific course, this exam is used as an admissions test for those applying to: maths, maths and philosophy, maths and stats, computer science, maths and computer science, computer science and philosophy. There are 7 questions but the applicants do 5 depending on what course they are applying for
@@jpnep8690 I'm seconding this question, educational system in my country is kinda trash.
For some reason it is still slightly stressful to see the MAT paper even though I am literally already in Oxford student
For part iii i think a better method would be to shift the function downward one half, then plug in -y for y and see if it yeilds the same equation.
Enjoyed the Question and ofcourse the video too. You should also have a look at the JEE advanced paper. It has some nice Questions in calculus and other math in general, could be good inspiration for your channel's audience!
Hi love ur vids, I had a request. Could you plz make some videos on the STEP Exam. It is the Cambridge entrance exam for High School students and the questions in it are absolutely brutal. There are 3 papers in total and they increase in difficulty from STEP 1 being the easiest and STEP 3 is the hardest. I think you will really enjoy some of the questions since in my opinion, having looked at both the JEE advance maths questions and STEP 3, I would say that a lot of the STEP 3 are actually even harder than Jee so plz give it a go. But if you want the really hard ones then do STEP 3. STEP 2 AND 1 are still hard but not as hard. Also, if you want the hardest ones, even from STEP 3, then there is a mark scheme which has all the answers of the questions from the past years and towards the end of that, you can go to STEP 3 and they tell you which questions were done well or poorly. Hence which were hardest and which were easiest.
STEP 1 got cancelled :(
Do you have a link?
@@blackpenredpen do you have an email, I can email you some cool STEP questions and stuff for you to have a go at!!
@@blackpenredpen Also a second note if you google “STEP support programme” and click the second link down you should see all the foundation modules they have many problems there.
@@Smevv Sure, blackpenredpen@gmail.com If possible, please include the solutions too. Thanks!
Wow! What an adventure! Nice work!
🤩
The MAT brings back some interesting memories - excited to watch this one!!
You should share your MAT experience. It will be interesting.
@@blackpenredpen I'm planning on doing this on my channel soon - how did you read my mind?! 😂 PS, great video - I can't think of a better way of spending Valentine's Day!
@J Pi Maths I subscribed. Looks like a good channel
@@tomatrix7525 thanks so much! Welcome to the Jπ family!
When I did this a few months ago I wasn’t able to do parts v and vi, the others ones I got correct but the last 2 where hard. I could’ve known v when I saw how it was done but vi was too hard for me
you have a copying mistake at 13:47
Yea... I just noticed that earlier during the live
I was going to point this out too but saw you already did.
I was not familiar with Vieta's formulas so the explanation for the last part of section vi left me very confused.
the missing piece was (-a sub n-1 / a sub n) = -1 / 1 = the sum of roots, where a sub i is the coefficient of the ith term in the cubic equation
great video! Finally somthing that is not related to lambert function or cimplex numbers
another way to solve (ii) is to find dx/dy and set it equal to 0. you get (2y-1)/(3x^3-1)=0 which means y=1/2
for part 3 you can say that if (a,b) is on the curve then (a, 1-b) is on the curve and the distance to the line y=1/2 from each is abs(b-1/2) without writing any equations. Once I wrote it it is probably not faster than what you did.
I went with the wording in the solution since I was not sure how strict their grading was.
set y = 0. x intecpt values∈{ -1,0,1} . due to the restrictions, the values are 1,-1.
set x = 0. y intercept values∈{ 0,1}
y is a parabola, since parabola's vortex is midway between the y-intercepts, the vortex of the graph is symmetric on the line y=0.5
since parabolas are symmetric on two sides of the vortex and the other side is independent of y, the graph is symmetric on the line y=0.5. (there are some circular reasoning that I'm not energized enough to resolve)
since the three numbers already exist on the same horizontal line y=0.5, replace y with 0.5 in the original equation and move everything to one side gives x^3-x+0.25=0. replace the 0 with a new y gives a cubic equation that has the correct roots.
the equation of the circle can be written as y=sqrt( (b-a)^2/4+(x-(b+a)/2)^2 )+0.5 which can be simplified to sqrt( - (x-a) (x-b) )+0.5
sub it into the "y" of the original problem can probably yield the correct answer. however, the amount of calculation already feels like this is the wrong path.
however, by continuing the next step, the roots actually cancel out nicely to 0 = x^{3}+x^{2}-(1+a+b)x+(ab+0.25). then I got stuck since I don't know the cubic roots formula. so maybe this is indeed the wrong path.
try again and define the circle as a function of y(right half) might work, but i really need to be doing my homework by now.
I have the link to the problems in the description. You can try the questions first and then watch the video tomorrow. : )
I think we can try finding ∆ (delta, couldn't find the other one) by setting the denominator of dy/dx to be zero
If this was in my secondary school, I felt like this could be doable but it would be quite hard
try this: elliptic integral of the 2nd kind, integrate sqrt(1+c*sin^2(z) ] dz = (2/3)*csc^2(z)*(c*sin^2(z)+1)^(3/2), the c-value is negative, csc has a zero-point issue
This was easier than I thought it was.....I was able to solve most of of except the 6th on my own....although in the exam hall scenario I would panicked and definitely skip it.....it was an excellent problem tho
For part 2, I had no idea for to solve it but to find the axis of symmetry, which is y=?, I just did -b/2a in terms of y.
This vid is going to be real good
Can u make some videos on Jee advanced limits questions pls?
Elliptic curves have an A+ on intimidation, let me tell you.
Really cool problem!!! Thank you for sharing
I once did a paper which purely consited of greek letters, and when I was doing it, I gave up because as I progressed, all of the letters just began to look identical
Kindly 😔 make a video on factorial sumes
Hey blackpenredpen! Can u solve the following integral: ∫(x³ - 4x² + 3x + 6)/√(x² + 3x + 5) dx evaluated from 0 to 4.
I challenge you to solve this without using any of the hyperbolic functions or their inverses... use the logarithm instead... By the way, great video as always!! I am a huge fan of yours!
That was a lot of fun. ¡¡¡Thank you !!!
Integral of tan e^senx/sec x
I did the Oxford maths entrance exam in 1993, and scored 98% on the pure paper. The funny thing was that loads of other people got very similar marks. They’d just made the paper a little too easy / similar to previous years. In contrast very strong candidates were getting in with very low scores (like 40%) the year before - the 1992 paper was just a bit too hard. The margins can be quite small at this level for a bunch of pretty well prepared candidates.
@blackpenredpen i am thirsty, not for water, not for integrals, not for even the most delightfull of elliptic curves. I am thirsty for some beautiful authentic bprp socks. thankyou for your content, i just must open my wallet at long last to purchase some math for fun socks. If you could drop some merch socks i would be so thankful kind sir :))
BPRP: Can you solve....?
Me: No, but it’s VERY satisfying to watch someone else solve it.
There is an equation that popped into my mind and I still don't know how to solve it.
Sin (x)=x+5
Of course, x needs to be negative because sin is only between -1 and 1 or maybe the answer is a complex number but I still don't know how to approach it. According to WolframAlpha, the answer is -4.15252 but I don't know how they got the answer I think they plotted both graphs and found the intersection point but how can we solve it algebraically. Please help
That's pretty fast sir you should upload this on bprpfast 😂😂😂
I think I did this one when I took it, the binomial numbers one was pretty hard I thought
Hey I remember answering this question, it was a pretty nice question tbh!
Btw i ended up geting 30 points out of 100 which is kinda crappy but hey i enjoyed the test and i appreciate the opportunity at least
Missed this guys videos for a bit and now he’s a viking
Gotta catch em all
Is there any way to know if a pub keys private key is odd or even regardless of range in secp256k1 ?
Quest 6: In terms of delta? SO where is delta?
I miswrote it. It should be in terms of gamma not delta.
Man... all the Greek letters in that question >_
I was pulling my hair out trying to solve it for delta XD
Please solve integration of [sqrt(x) sinx] dx
Happy Valentine’s Steve!!! Nice video as always
At 2:56, I used implicit differentiation to solve for the y-value. I prefer to, and it is easier for me lol.
Differentiate both sides with respect to y and also got the answer δ=0.5.
(Although a bit more working out)
Also at 4:01,
dx/dy=(2y-1)/(3x^2-1);
Let dx/dy=0 (vertical gradient), x can be any Real Number (except x=±1/√3) and y is always 1/2.
For gamma since the curve is symmetric across the value of gamma, I plug in b for x. Then I solve for both y's. The average of those y's is gamma. And you get gamma=1/2.
I did this and nailed it :D
It was a fun question. Not sure if I can solve @/vi directly without those previous small steps.
For the sixth directly use the diametric equation of a circle
Can you guys solve this, this is from peterson related rates problem
one ship is sailing south at a rate of 5 knots, and another ship is sailing east at a rate of 10 knots. at 2 P. M. the second ship was at the place occupied by the first ship one hour before. at what time was the distance between the ships not changing?
My take on part 2 with some calculus:
Let g : (-1,1) -> R^2 be a one to-one differentiable path with the image of g completely within the given curve, furthermore suppose that g' is never zero, and suppose g(0)=(x_0,δ) where x_0 is one of α,β,γ.
Note: since F(x,y)=x^3-x-y^2+y is continuously differentiable and ∇F(x,y)=(3x^2-1, -2y+1), we have that the conditions for the implicit function theorem (for extraction of one of the variables as a function of the second) hold everywhere except for the points where y=1/2, x=±√1/3, in particular those points are not on the curve, and so such a path does indeed exist for any point on the curve that we choose.
Now after using a paragraph to justify the existence of g, we know by geometric arguments that at the point g(0) we have that g'(0)=(0,y'(0)) for some nonzero y'(0), but writing g(t)=(x(t),y(t)) and differentiating the curve's equation (which holds for the image of g):
(2y(t)-1)y'(t)=(3x(t)^2-1)x'(t)
and in particular
(2y(0)-1)y'(0)=0
and since y'(0) is nonzero, we have y(0)=1/2, so δ=1/2
What is answer of int 0 to 2pi. {(acost)^2+(-bsint)^2} ^1/2 dt
Where a= 0.3870 b=0.3788
Can anyone help me out?
I'm just guessing at this point (only 2.38 into the video) but y=delta is a line of symmetry and on the lhs we have y^2-y=y*(y-1) so there is a symmetry mid way between y=0 and y=1. (because if you think about it (swapping y and y-1 would leave the equation unchanged)) (lol not only was i right but answered the next question as well.)
On iii, isn't it enough to notice that the LHS is even w.r.t. y-1/2 = y-δ, and the RHS is independent of y (and therefore y-1/2 or y=δ)?
can you make an entire playlist for calc 3 out of ur videeos
Isn't it much easier for proofing the symmetry to replace y with 1/2 - (y - 1/2)? Simplify to 1 - y. f(x, y) := y^2 - y = x^3 - x. Then f(x, 1 - y) is (1 - y)^2 - (1 - y) = x^2 - x => 1 - 2y + y^2 - 1 + y x^2 - x => y^2 - y = x^3 - x
Hi blackpenredpen, could you make a video explaining how to solve
\sqrt [x]{x} = 1.35
13:11
All we have to do is
Stay a minute!!!
where can i find that laplace transform poster?
It’s from my Teespring store. The link is in the description 😃
@@blackpenredpen thank you! love your videos!
Parts 1 2 and 3 were a lot easier than I first expected, you just have to think about it a bit. I am very surprised
I think I have a cleaner explanation of why the curve is symmetric about the line. For any point (X,Y) on the curve we know Y^2 - Y = X^2 - X. Now consider the point (X,Y') which is (X,Y)'s reflection about the line y = 1/2. We have (Y + Y')/2 = 1/2 since the midpoint must be on the line, so Y' = 1 - Y. Therefore Y'^2 - Y' = (1 - Y)^2 - (1- Y) = 1 - 2Y + Y^2 - 1 + Y = Y^2 - Y = X^2 - X. So the point (X,Y') also lies on the curve and hence the curve is symmetric about y = 1/2.
Where do you find your ideas for the videos, man?
It s so cool your job
this is random but how would you find the area of the little circle/oval/ellipsoid thingy on the left? y cant be defined explicitly in terms of x in a nice way, and i dont think polar coordinates would work here
Hello, can you make a video about the prime zeta function?
3:25 why is y=0.5?
We want alot of this problems please😅
I NEED TRIG SHIRTS AND HOODIE!!!
I can't believe such an easy was asked
Cambridge offer holder here, and I couldn't solve this! Only got as far as part ii
Ono I did this question in the actual exam and messed it up
12:10 why isn't is (alpha + beta)/2
Because you are finding the distance from beta to alpha. To do this you minus alpha from beta. Then divide by two for radius. Like if you had 2 and -1. The distance would be 2-(-1)= 3 . Alpha + beta / 2 finds the arithmetic average between the two points with corresponds to a number on the x axis which is the distance from the y axis.
Evaluate integral from zero to infinity xdx/ 1+e^x challenging question
There are another solutions for log_i(i) other than 1?
In (iii) wouldn’t be the point symmetrical to (x,1/2sqrY) (x,-1/2-sqrY) instead of (x,1/2-sqrY)?
If the quadratic equation can be solved use the formula, namely the formula abc. can cubic equations also be solved using formulas ?, if you can use what formulas?
Can you make some video on vector and 3 D
Why is my math soooo rigorous even in high school? Like soooo many epsilon delta proofs (we use alpha and beta instead of epsilon and delta xD) and many techniches for reasoning, not to mention the EXTREMELY mind bending geometry, our math curriculum is originally french, so maybe that's why.
I make calculus videos please check them out and subscribe
French people really prepare for things like Abstract Algebra and pure mathematics. Not sure if it's because of Galois.
Can you solve this:
log_6^x=log_3^4x
I really need it, thanks!!!
Can you calculate the perimeter of a ellipse?
Hey why dont u try some IIT JEE ADVANCED Qs....it will fun to see u solve them
I Think You Should Attempt JEE Advanced Maths Section and GATE Maths That Would Be Fun As JEE ADVANCED Is For High School Student
Even this is for high school students
Can you please do IIT JEE Advanced 2020 paper analysis. 🙏🏻🙏🏻
14:20 the signs from 2nd eqn are messed up
Why not use a change of variables in part 2 and reason about symmetry about y=0 instead?
I make calculus vids please check them out and subscribe if you like my content
Why do you choose add 1/4 ? Please explaind me i don't understand that
He's using (ay + b)^2 = (a^2)(y^2) + 2aby + b^2 in reverse, picking values of a and b that will end up as (y^2 - y) plus some constant. So a^2 must be 1 and 2ab must be -1. Choose a = 1 and b = -1/2 to satisfy this, so b^2 is 1/4. Then observe that (y - 1/2)^2 = (y^2 - y + 1/4) according to the formula I gave in the beginning, so that's what you want to achieve on the right-hand side.
can you help me find the integral of square root of sinhx
Thank you
need your help man. Integrale of x/ln(x)dx ? thx so mutch for your help :)
Hi plz solve my problem 0 to π ∫ dx/[ (5+4cosx) ^2 ] =???
Shouldn't the last answer be in terms of delta?
*I have a very serious doubt*
*Why is this happening*
*differentiation of sin²x = sin 2x*
*And*
*Integration of sin 2x is -cos 2x /2*
Please address it
If I am right, it is right to say that if we integrate a function and then differentiate the answer from integration , we will get the function back ?
you should have used ξ we accually use it more often in greece and it would have made your life harder
somehow i saw how first you wrote: x^3-x+1/4 and then x^3+x-1/4. You switched the sign by accident. Ow
hello, 1^x=-2 is solvable?
hi bro can usolution this for me
1/2+1/3+1/4+.....+1/n+1
Try JEE ADVANCED Maths Lets see what do you think about it
Could implicit differentiation also have worked?
That's pretty interesting
I make calc videos please check them out and subscribe if you like my content
台灣時間半夜1點直播
喔那會折磨死我
我應該是無法撐到那麽晚不睡
喔, 這不是直播, 這只是我也會在線上跟你們一起看. 影片之後也會還在 : )
Oh yeah
Which language are you speaking please??
@@Hicham_saidi mandarin
100 everything on conic section PLEASEEEEE.🙂🙂