Do you want to understand the most famous Euler number e like NEVER before? 🎉 ua-cam.com/video/fZTfEvM7Wys/v-deo.htmlsi=KSIZnicTf9cqNe0j Check out my short video to learn its true meaning that most mathematicians don't know but ANYONE can understand! 🥳
Thank you. The derivation is (as is often the case) easier to remember (x=u+v) than the resulting formula. And: Now I understand for the first time why the solution has this form. Great!
Hi @@WolfgangFeist ! Thank you so much for your comment! Yes, I agree, I wouldn't want to memorize the formula for solving the cubic (especially in non-depressed form). I get the sense that part of the reason it isn't widely taught is because the formula (at least the non-depressed version) is extremely complicated, but the approach is really beautiful and creative! I'm so happy I was able to share it with you! 😊 (A question that is also interesting is: what is the intuition behind the substitution x = u + v? Yes, it works, but is there deeper reasoning for it (or does it fit into a broader context where such substitutions solve problems/give insights)? I don't know if I have a satisfactory answer for this personally, although I thought about it a bit.) I hope you have an amazing day/evening/night! 😊 I also wish you a very happy holiday season and amazing New Year for you in 2025! 🥳 P.S. I didn't forget to reply to your questions on the other video, I apologize for the delay and I will do so! 😊
Motivation for x = u + v? Idea: the cubic polynomial -q is an odd function in the depressed case: So symmetric to x=0, y=q. The extrema are therefore symmetric to x=0; in the case q=0, so are potential real solutions (another is 0). If we now shift the graph by q... :-)??
Better idea: Let's assume p(x) = (x-x1)(x-x2)(x-x3); for the depressed cubic equation, the coefficient a2 of x² is a2=0. Comparing the coefficients, we get x1 + x2 + x3 = -a2 = 0; implication: x1 = -x2 - x3; let's call u = -x2, v = -x3 and that's exactly the "trick"; which incidentally tells us that x2=-u and x3=-v are also solutions! Nice. So the whole thing is really straightforward. (If you know about it... :-)
That motivation: OK. But: the part with the 'incidentally ... x2=-u" is not true (in general), I was too fast in such a conclusion. x1 can be a sum of OTHER u1 and v1, which are NOT roots but satisfy that u³+v³=q and 3uv=p.
Excellent. Amitesh does not dumb down or slow down. However, the small whiteboard means he is constantly wiping formulae, so the listener loses the thread. A modern electronic whiteboard is needed, which revolutionizes lecturing into the 21st century.
Hi @cosimo7770 thanks so much for your comment and for the feedback! 😊 I really appreciate it! I agree a small whiteboard isn't ideal. I have never used an electronic whiteboard but I guess what it would do is allow me to go back to previous "slides" before they were erased. At present, however, at least I think that someone can backtrack through a video to recall any points and that might be instructive anyway to help them connect the points in their own mind (again, not ideal, but the information at least isn't lost). I'm still a small UA-camr, but I hope that as my channel grows, I can invest more into a sophisticated video production! I would like to have my own studio and have lots of ideas which I would like to implement in general! However, at the beginning of my UA-cam journey it was cool to see many UA-camrs with a million+ subscribers (even in the math sphere) who don't do anything fancy editing-wise or production-wise, just explain math at a whiteboard. 😊 I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
Hi @onafehts thank you so much for your kind comment and for sharing! 😊 I will have to check that video out, it sounds really interesting! 😊 I am glad to hear that the UA-cam algorithm is working well 😅, and it's great to get that feedback from you. I'm excited to have you as a subscriber and I am looking forward to producing more content you will enjoy! I wish you a very Happy New Year!!! 🥳🎉🎊
Do you want to understand the most famous Euler number e like NEVER before? 🎉 ua-cam.com/video/fZTfEvM7Wys/v-deo.htmlsi=KSIZnicTf9cqNe0j Check out my short video to learn its true meaning that most mathematicians don't know but ANYONE can understand! 🥳
Thank you. The derivation is (as is often the case) easier to remember (x=u+v) than the resulting formula. And: Now I understand for the first time why the solution has this form. Great!
Hi @@WolfgangFeist ! Thank you so much for your comment! Yes, I agree, I wouldn't want to memorize the formula for solving the cubic (especially in non-depressed form). I get the sense that part of the reason it isn't widely taught is because the formula (at least the non-depressed version) is extremely complicated, but the approach is really beautiful and creative! I'm so happy I was able to share it with you! 😊
(A question that is also interesting is: what is the intuition behind the substitution x = u + v? Yes, it works, but is there deeper reasoning for it (or does it fit into a broader context where such substitutions solve problems/give insights)? I don't know if I have a satisfactory answer for this personally, although I thought about it a bit.)
I hope you have an amazing day/evening/night! 😊 I also wish you a very happy holiday season and amazing New Year for you in 2025! 🥳
P.S. I didn't forget to reply to your questions on the other video, I apologize for the delay and I will do so! 😊
Motivation for x = u + v? Idea: the cubic polynomial -q is an odd function in the depressed case: So symmetric to x=0, y=q. The extrema are therefore symmetric to x=0; in the case q=0, so are potential real solutions (another is 0). If we now shift the graph by q... :-)??
Better idea: Let's assume p(x) = (x-x1)(x-x2)(x-x3); for the depressed cubic equation, the coefficient a2 of x² is a2=0. Comparing the coefficients, we get x1 + x2 + x3 = -a2 = 0; implication: x1 = -x2 - x3; let's call u = -x2, v = -x3 and that's exactly the "trick"; which incidentally tells us that x2=-u and x3=-v are also solutions!
Nice. So the whole thing is really straightforward. (If you know about it... :-)
That motivation: OK. But: the part with the 'incidentally ... x2=-u" is not true (in general), I was too fast in such a conclusion. x1 can be a sum of OTHER u1 and v1, which are NOT roots but satisfy that u³+v³=q and 3uv=p.
Excellent. Amitesh does not dumb down or slow down. However, the small whiteboard means he is constantly wiping formulae, so the listener loses the thread. A modern electronic whiteboard is needed, which revolutionizes lecturing into the 21st century.
Hi @cosimo7770 thanks so much for your comment and for the feedback! 😊 I really appreciate it! I agree a small whiteboard isn't ideal. I have never used an electronic whiteboard but I guess what it would do is allow me to go back to previous "slides" before they were erased. At present, however, at least I think that someone can backtrack through a video to recall any points and that might be instructive anyway to help them connect the points in their own mind (again, not ideal, but the information at least isn't lost).
I'm still a small UA-camr, but I hope that as my channel grows, I can invest more into a sophisticated video production! I would like to have my own studio and have lots of ideas which I would like to implement in general! However, at the beginning of my UA-cam journey it was cool to see many UA-camrs with a million+ subscribers (even in the math sphere) who don't do anything fancy editing-wise or production-wise, just explain math at a whiteboard. 😊
I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
Nice job, man! Subscribed. I watched a video on Veritassium about the story of that discovery and I think that was why your video showed up to me.
Hi @onafehts thank you so much for your kind comment and for sharing! 😊 I will have to check that video out, it sounds really interesting! 😊 I am glad to hear that the UA-cam algorithm is working well 😅, and it's great to get that feedback from you. I'm excited to have you as a subscriber and I am looking forward to producing more content you will enjoy! I wish you a very Happy New Year!!! 🥳🎉🎊