Hi, great video this is very similar to my Further Engineering Maths book I wonder if this is a standard way of teaching the subject? though more seriously how do you deal with the infinite cases of these polygons, I have been shown the "triangle" with a single vertex being taken to infinity however I don't see how that limiting value for the term in the product will go to 1 i.e. (w-u)^{\frac{α}{\pi}-1} I would assume as you take the vertex to infinity the α would go to 0 thus the exponent would go to -1 thus you would get (w-u)^-1 which isn't the expected behaviour of at the limit the value of this term in the product going to 1. Is there a better way of explaining how this limiting process works?
Hi there. Appreciate the engagement. Part of my approach is inspired by the book Advanced Engineering Mathematics by Stroud. It's quite intuitive, and I haven't seen it in many other books. I hope I've understood your query correctly. You are right that the exponent goes to -1. This means that u goes to infinity, and the term behaves like 1/u. Taking the limit suggests that the term approaches 0 rather than 1. However, the overall integral doesn't go to 0. This terms "smallness" is compensated by the rest of the integrand and the overall scaling factor "A".
@@SeriousSolvers yh that makes sense my book is further engineering maths by KA Stroud so that probably the similarity (Same Author). thanks and for the reply on my other question, I take it that as u goes to infinity the w in the denominator becomes negligable fast enough that as you take the limit the term becomes like a constant "sooner" that it becomes like zero
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on some of bounded conditions, by mistake U forgot signed. For ex., hwf can it be: infinity
Excellent!
Thank you, glad you liked it!
Hi, great video this is very similar to my Further Engineering Maths book I wonder if this is a standard way of teaching the subject? though more seriously how do you deal with the infinite cases of these polygons, I have been shown the "triangle" with a single vertex being taken to infinity however I don't see how that limiting value for the term in the product will go to 1 i.e. (w-u)^{\frac{α}{\pi}-1} I would assume as you take the vertex to infinity the α would go to 0 thus the exponent would go to -1 thus you would get (w-u)^-1 which isn't the expected behaviour of at the limit the value of this term in the product going to 1. Is there a better way of explaining how this limiting process works?
also fyi I'm annoyed that I didn't find this channel earlier this is nearly exactly what I have been looking for on yt thx keep up the good work
Hi there. Appreciate the engagement. Part of my approach is inspired by the book Advanced Engineering Mathematics by Stroud. It's quite intuitive, and I haven't seen it in many other books.
I hope I've understood your query correctly. You are right that the exponent goes to -1. This means that u goes to infinity, and the term behaves like 1/u. Taking the limit suggests that the term approaches 0 rather than 1. However, the overall integral doesn't go to 0. This terms "smallness" is compensated by the rest of the integrand and the overall scaling factor "A".
Thank you! Glad it's helpful
@@SeriousSolvers yh that makes sense my book is further engineering maths by KA Stroud so that probably the similarity (Same Author). thanks and for the reply on my other question, I take it that as u goes to infinity the w in the denominator becomes negligable fast enough that as you take the limit the term becomes like a constant "sooner" that it becomes like zero