Yes, correct. What I show in the video is only half the solution. At the end, the final solution is the linear combination (i.e. added with coefficients) of the two solutions corresponding to each r value.
So, if we disregard the xʳ part, and focus only on the power-series part, what would be the differential equation that gives just that part, and how would it be related to the original equation?
Try plugging in x^r*f(x) into the original diff eq. You will see that the d.e. for f still has a singularity (since you only get 1 frobenius solution for each value of r.)
So what is that function in the parenthesis? y(x) = a₀ · √x · f(x) where: f(x) = 1 + x²/(2·7) + x⁴/(2·7)/(4·11) + x⁶/(2·7)/(4·11)/(6·15) + x⁸/(2·7)/(4·11)/(6·15)/(8·19) + … I can see some factorial-like stuff going on in there in the denominator: f(x) = 1 + x²/(2·7) + x⁴/(2·4)/(7·11) + x⁶/(2·4·6)/(7·11·15) + x⁸/(2·4·6·8)(7·11·15·19) + … so it seems to contain the products of even numbers. Can it be rewritten in terms of factorials? Hmm... But how can we make factorial of only _even_ numbers and exclude the odd numbers from the product?.. :|
Your videos contain the best explanations for the Frobenius Method. Thank you.
Brilliant explanation✅ I finally understand this method.
Thank you sooo much sir ! You did answers all my mysterious wizardly questions i had before 😂❤
Extremely helpful series, taking math methods rn.
Thank you so much! Wished I have you as my lecturer
Thank you! Its very helpful!!
you helped me a lot! Thank you!
Obrigado professor!
You're the best!
Your students are so lucky!
Very nice video
Good explanation ❤...
And I have a doubt, In some other videos they added the answers from the both r values to the final answer. Is that correct?
Yes, correct. What I show in the video is only half the solution. At the end, the final solution is the linear combination (i.e. added with coefficients) of the two solutions corresponding to each r value.
So, if we disregard the xʳ part, and focus only on the power-series part, what would be the differential equation that gives just that part, and how would it be related to the original equation?
Try plugging in x^r*f(x) into the original diff eq. You will see that the d.e. for f still has a singularity (since you only get 1 frobenius solution for each value of r.)
So what is that function in the parenthesis? y(x) = a₀ · √x · f(x) where:
f(x) = 1 + x²/(2·7) + x⁴/(2·7)/(4·11) + x⁶/(2·7)/(4·11)/(6·15) + x⁸/(2·7)/(4·11)/(6·15)/(8·19) + …
I can see some factorial-like stuff going on in there in the denominator:
f(x) = 1 + x²/(2·7) + x⁴/(2·4)/(7·11) + x⁶/(2·4·6)/(7·11·15) + x⁸/(2·4·6·8)(7·11·15·19) + …
so it seems to contain the products of even numbers. Can it be rewritten in terms of factorials?
Hmm... But how can we make factorial of only _even_ numbers and exclude the odd numbers from the product?.. :|
Thank you so much! :D
Thankyou bro
7:13 scared the s** in me hahhahaha
Than q sirr
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