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confrac,matrix function 2.

lvl107
20-Turquoise

confrac,matrix function 2.

Hello, Everyone.

confrac%2Cmatrix+function+2.PNG

Thanks in advance for your time and help.

Best Regards.

ACCEPTED SOLUTION

Accepted Solutions
Werner_E
25-Diamond I
(To:lvl107)

I don't know why you are confused. What you are doing the whole time is calculating approximations. With different substitutions you get different aproximation-functions. This is especially true as you compare a linear substitution to a nonlinear one. Your 1/x substitution has quite some advantages, though.

While I can see no reason why someone would do that I find it strange and have no suitable explanation for tha fact that applying confrac a second time makes the approximation that bad.

confrac5.xmcdz.png

View solution in original post

5 REPLIES 5
Werner_E
25-Diamond I
(To:lvl107)

confrac3.png

Werner_E
25-Diamond I
(To:lvl107)

As with a Taylor series you can have a continued fraction, but not a standard one with respect to x but e.g. with respect to x-1

confrac4.png

lvl107
20-Turquoise
(To:Werner_E)

Thanks for your response, Werner.

confrac%2Cmatrix+2.PNG

Best Regards.

lvl107
20-Turquoise
(To:lvl107)

or:

confrac%2Cmatrix+2.PNG

Werner_E
25-Diamond I
(To:lvl107)

I don't know why you are confused. What you are doing the whole time is calculating approximations. With different substitutions you get different aproximation-functions. This is especially true as you compare a linear substitution to a nonlinear one. Your 1/x substitution has quite some advantages, though.

While I can see no reason why someone would do that I find it strange and have no suitable explanation for tha fact that applying confrac a second time makes the approximation that bad.

confrac5.xmcdz.png

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