Query on indexing to determine coefficients

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The discussion centers on the validity of the expression for calculating coefficients in the equation B_ij, particularly when i=1 and j=1. Participants note that the author claims the expression is invalid for these values, yet calculations yield specific results for B_11 and B_1j. Confusion arises over the indeterminate forms encountered in the equation, especially regarding the second term in the brackets, which results in 0/0 when i=j=1. There is also mention of potential editing issues in the expression that could affect clarity. The conversation seeks clarification on how these mathematical nuances impact the overall validity of the calculations.
bugatti79
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Folks,
I am interested to know what the author is doing in the following

##\displaystyle B_{ij}=EL ij (L)^{(i+j-1)} \left[ \frac{(i-1)(j-1)}{i+j-3} -\frac{2(ij-1)}{i+j-2}+\frac{(i+1)(j+1)}{i+j-1}\right]##

he states that this expression is not valid for ##B_{ij}## when ##i=1## and ##j=1,2,...N##

...yet he goes on to actually calculate

##B_{11}=4EIL##, ##B_{1j}=B_{j1}=2EIL^j##, ##(j>1)##

I understand the the numerator in the first 2 terms inside the big brakets are both 0 when i=j=1 but we still yield a value from the third term...
Any insight will be appreciated
Regards

PS:I notice there is some editing problem with the 3 terms inside the big brackets. There should be a minus and plus separating the terms.
 
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bugatti79 said:
Folks,
I am interested to know what the author is doing in the following

##\displaystyle B_{ij}=EL ij (L)^{(i+j-1)} \left[ \frac{(i-1)(j-1)}{i+j-3} -\frac{2(ij-1)}{i+j-2}+\frac{(i+1)(j+1)}{i+j-1}\right]##

he states that this expression is not valid for ##B_{ij}## when ##i=1## and ##j=1,2,...N##
That's not at all obvious. The only restrictions I see are that
1. i + j ≠ 3 (would make the first denominator vanish)
2. i + j ≠ 2 (would make the second denominator vanish)
3. i + j ≠ 1 (would make the third denominator vanish)
bugatti79 said:
...yet he goes on to actually calculate

##B_{11}=4EIL##, ##B_{1j}=B_{j1}=2EIL^j##, ##(j>1)##
I don't see how. With i = 1, j = 1, the second term in the brackets is 0/0.
bugatti79 said:
I understand the the numerator in the first 2 terms inside the big brakets are both 0 when i=j=1 but we still yield a value from the third term...
Any insight will be appreciated
Regards

PS:I notice there is some editing problem with the 3 terms inside the big brackets. There should be a minus and plus separating the terms.
 
Could you give a link to where you found this question?
 
Mark44 said:
I don't see how. With i = 1, j = 1, the second term in the brackets is 0/0.

Are you saying because one of the terms is indeterminate then the whole equation is invalid and thus cannot be usedt o calcualte ##B_{ij}## for i=j=1?

micromass said:
Could you give a link to where you found this question?

See attached jpeg of question. The answer involves converting the DE into a weak form using a weight function w and splitting the differentiation between the weight function and the dependent variable u.
Would the choice of the approximation functions affect the outcome?

regards
 

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Good morning I have been refreshing my memory about Leibniz differentiation of integrals and found some useful videos from digital-university.org on YouTube. Although the audio quality is poor and the speaker proceeds a bit slowly, the explanations and processes are clear. However, it seems that one video in the Leibniz rule series is missing. While the videos are still present on YouTube, the referring website no longer exists but is preserved on the internet archive...

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