Recent content by Rochefort

I wanted you to prove that I could make ##A## the subject for when ##n=1## since I thought that you suggested that it was possible when using trig substitution and it didn't seem so. But you later specified Which I gave up on since I found it difficult to find my "order of approximation."...

I am trying to find a method that gives me values of ##A## given values of ##B## and ##C##.

Using Numerical Methods I agree, using a numerical method would be easier. Here is my method: Using I can find ##k## as a root For example, when ##B=2## and ##C=2.17363253251301 ## (I know ##A=1## using my Riemann sum) $$\frac{2E(1,k)}{\sqrt{1 - k^2}}- 2.17363253251301 = \dfrac {2\int...

I have followed your method up to this point. I am stuck on what you have told me to do here.

Yes, if you can make "A" the subject please do!

By solvable I assume you mean you can make "A" the subject. If you can prove this for n=1 I can finish my investigation (without using elliptical integrals). So please could you prove this?

Sorry, accidentally posted via my phone, my question is bellow.

@pasmith: That is way beyond my understanding (at the moment.) Could you provide more steps/explanation to your method. Also if n=10 for C=\int^{B}_{0}\sqrt{1+\left(\dfrac{A}{B}\right)^{2}+\sum^{n}_{k=1} \dfrac{A^{2}\left(-1\right)^{k}2^{2k-1}x^{2k}}{\left(2k\right)!B^{2k+2}}}dx could you show...

Also I don't think you can solve for "A"

I am finding it difficult to find a good approximation. I have tried different centers, "a" and different values for "A" and "B". Every time I get the summation up to the 6th order (when n=6) my computer screen freezes. Should I just assume "n" is a high value e.g. n=10? Could you suggest a way...

One last question. How do I find which values of "a" and "n" are the most appropriate?

Thank you so much. You have been so helpful!

I have never used a Taylor series before so I don't know what you mean exactly. Here is my attempt so far: So from knowing the power series representation of \cos\left( x\right)\cos^{2}\left( x\right) =1+\sum^{\infty }_{n=1}\dfrac {\left(-1\right)^{n}2^{2n-1}x^{2n}}{\left( 2n\right) !}...

Well I know that C=\int^{B}_{0}\sqrt {1+\dfrac {A^{2}}{B^{2}}\cos^{2}\left(\dfrac {x}{B}\right) }dx And when provided with the values of A and B I can simulate results for C. What I did was, I converted it into a Riemann sum and produced an excel spreadsheet (consisting of exactly 1048574 rows...

No, I don't know what that is. This is a part of my own investigation and a method for solving this would help me progress. I don't know the exact problem statement. All I can do is produce more equations of different values (on the LHS), for values of A. If you want, I can post more equations...