Thank you brother for your comment. If you have any particular question I will be happy to answer. Good luck with your studies, Electrical Engineering is awesome field.
Absolutely brother, the argument is valid for a second degree polynomial. That is absolutely correct. Both in the discrete and integral case we introduce polynomials of degree 2 and we employ the Discriminant to obtain the desired inequality. You can go again through the video if you wish, and if you have further questions feel free to ask. If you like the content of the channel feel free to subscribe, I am constantly uploading material and videos. Thanks for the comment my friend.
Thank you for your question. When you define the function P of lambda, you can see that this function is positive. When you expand the identity and split the integrals you see that you have a polynomial of variable lambda of degree 2 . When the discriminant is strictly negative, then you don't have real roots and this means that the polynomial is positive. If the discriminant is 0 then you have a double root which means that the polynomial is positive but becomes zero for this particular root. As a result we are interested in the case where the discriminant is less or equal to 0. Here is what you should take from the video: 1. Define the function P, 2. The function P >=0, 3. The function P is a polynomial of degree 2 and since it is >=0, the discriminant must be
There is a latest video with two applications of the Cauchy Schwarz inequality. Feel free to have a look my friend and let me know if something is unclear.
Hello, thank you a lot for your comment. You notice that when the Discriminant Delta is computed, there is the integral of the product of the 2 functions at the left hand side. Similarly, at the right hand side the product of two integrals. At some point you take the square root and after one small step the Cauchy Schwarz inequality arises naturally. Feel free to watch the video again and have a look at the Discriminant part. I hope this helps, let me know if you have another question.
@@alexandroskyriakis3675comment very thank you! But I don’t understand the process from the second to the last line. In detail, How did you know |integral(fg)| greater than integral(|f||g|)?
I have made a video ua-cam.com/video/bUTFjsv2SOA/v-deo.html proving the inequality that you have asked me before. Feel free to have a look, this will make things more clear. Have a great day.
Please, can you explain why by applying | integral f g | < integral |f| |g|, the last inequality is necessarily valid (the final form of the Cauchy Schwarz inequality) ?
@@EPrevost Thank you for your comment, very accurate and it means you thought deeply about the last bound. I have employed triangle type inequality for integral. You take that - |fg|
@@alexandroskyriakis3675 Thank you very much for your help ! I understand the video about the triangle inequality for integral. Maybe it's a dumb question but, by applying it to the Cauchy Schwarz inequality, how can we be sure that integral |f| |g| is still less or equal to the right hand side (product of the two integral) ? To simplify, if a ≤ c and a ≤ b, how can we prove that b ≤ c ? Thank you in advance for your answer !
@EPrevost I have made a video with explanation, check it out. There are no dump questions, feel free to ask for more. Everybody struggles with maths, and if you know anybody who is perfect at maths please introduce this person to me. Thank you for your comments, much appreciation.
Sir thanks very 🎉🎉🎉🎉.
I enjoyed it. Please keep uploading you will be our teacher.
Electrical engineer in Cameroon🇨🇲🇨🇲
Thank you brother for your comment. If you have any particular question I will be happy to answer. Good luck with your studies, Electrical Engineering is awesome field.
What happens if the integral of f^2 is zero? I think that argument you used preveously is correct only for a polynomial of degree 2
Absolutely brother, the argument is valid for a second degree polynomial. That is absolutely correct. Both in the discrete and integral case we introduce polynomials of degree 2 and we employ the Discriminant to obtain the desired inequality. You can go again through the video if you wish, and if you have further questions feel free to ask. If you like the content of the channel feel free to subscribe, I am constantly uploading material and videos. Thanks for the comment my friend.
thank you
Please, explain why the discriminant Δ IS not greater than 0?
Thank you for your question. When you define the function P of lambda, you can see that this function is positive. When you expand the identity and split the integrals you see that you have a polynomial of variable lambda of degree 2 . When the discriminant is strictly negative, then you don't have real roots and this means that the polynomial is positive. If the discriminant is 0 then you have a double root which means that the polynomial is positive but becomes zero for this particular root. As a result we are interested in the case where the discriminant is less or equal to 0.
Here is what you should take from the video:
1. Define the function P,
2. The function P >=0,
3. The function P is a polynomial of degree 2 and since it is >=0, the discriminant must be
Can you make an example of its application?
There is a latest video with two applications of the Cauchy Schwarz inequality. Feel free to have a look my friend and let me know if something is unclear.
@@alexandroskyriakis3675 thanks!
I don't understand last process. Why integral [ absolute(f)*absolute(g) ] less than [integral f^2]^(1/2)[integralg^2]^(1/2) ??
Hello, thank you a lot for your comment. You notice that when the Discriminant Delta is computed, there is the integral of the product of the 2 functions at the left hand side. Similarly, at the right hand side the product of two integrals. At some point you take the square root and after one small step the Cauchy Schwarz inequality arises naturally. Feel free to watch the video again and have a look at the Discriminant part. I hope this helps, let me know if you have another question.
@@alexandroskyriakis3675comment very thank you! But I don’t understand the process from the second to the last line. In detail, How did you know
|integral(fg)| greater than integral(|f||g|)?
Thank you for the comment, much appreciation. The | integral f g|
@@alexandroskyriakis3675
Understood well! Thank you so much!
I have made a video ua-cam.com/video/bUTFjsv2SOA/v-deo.html proving the inequality that you have asked me before. Feel free to have a look, this will make things more clear. Have a great day.
every clear,and who can think of that constructor function like that to prove that,omg,but anyway,thank you
Thank you 4 the video
But please remember that
abs(int(fg)) =
I have done a video on this inequality. Feel free to have a look. Thank you again for this wonderful comment.
Please, can you explain why by applying | integral f g | < integral |f| |g|, the last inequality is necessarily valid (the final form of the Cauchy Schwarz inequality) ?
@@EPrevost Thank you for your comment, very accurate and it means you thought deeply about the last bound. I have employed triangle type inequality for integral. You take that - |fg|
@@alexandroskyriakis3675 Thank you very much for your help ! I understand the video about the triangle inequality for integral.
Maybe it's a dumb question but, by applying it to the Cauchy Schwarz inequality, how can we be sure that integral |f| |g| is still less or equal to the right hand side (product of the two integral) ? To simplify, if a ≤ c and a ≤ b, how can we prove that b ≤ c ?
Thank you in advance for your answer !
@EPrevost I have made a video with explanation, check it out. There are no dump questions, feel free to ask for more. Everybody struggles with maths, and if you know anybody who is perfect at maths please introduce this person to me. Thank you for your comments, much appreciation.
@@alexandroskyriakis3675 Thank you so much for your time, it helped me !
sigma male
Thank you. I don't have a clue what that means but I take it as a compliment.