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Original Message (ID '467792') By Alexander:
This is it, thank you! (I can now clearly see that this formula cannot be used to approximate the derivative at b[n] as the coefficients do not converge backwards.) Just one further question: In my formula I have the sum signs instead of Sum[] and I'm getting: Out[2]= Symbol/6 - (6*b[1])/5 + (15*b[2])/4 - (20*b[3])/3 + (15*b[4])/2 - 6*b[5] + (49*b[6])/20 What does that 'Symbol' mean?