Nov 8, 2006 #1 T tomshack Junior Member level 3 Joined Mar 4, 2006 Messages 27 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,286 Activity points 1,474 what is the taylor series expansion of this ? 1 ÷ (1+(wb)²)
Nov 8, 2006 #2 pmonon Full Member level 3 Joined Jan 28, 2006 Messages 152 Helped 7 Reputation 14 Reaction score 1 Trophy points 1,298 Activity points 2,232 which one is the indep. variable, w or b?
Nov 8, 2006 #3 T tomshack Junior Member level 3 Joined Mar 4, 2006 Messages 27 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,286 Activity points 1,474 pmonon said: which one is the indep. variable, w or b? Click to expand... b
Nov 8, 2006 #4 E-design Advanced Member level 5 Joined Jun 1, 2002 Messages 2,014 Helped 507 Reputation 1,016 Reaction score 408 Trophy points 1,363 Activity points 14,193 1+b².w²+[(w.b)²]²
Nov 8, 2006 #5 T tomshack Junior Member level 3 Joined Mar 4, 2006 Messages 27 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,286 Activity points 1,474 E-design said: 1+b².w²+[(w.b)²]² Click to expand... i think this is not the right answer.........
Nov 8, 2006 #6 pmonon Full Member level 3 Joined Jan 28, 2006 Messages 152 Helped 7 Reputation 14 Reaction score 1 Trophy points 1,298 Activity points 2,232 if b is the indep. variable and w is const., 1/(1+(wb)2) = 1/(1+(wδ)2) - {2wδ/(1+(wδ)2 )-2}(b-δ) - ........ + ….. see the attachment for further...
if b is the indep. variable and w is const., 1/(1+(wb)2) = 1/(1+(wδ)2) - {2wδ/(1+(wδ)2 )-2}(b-δ) - ........ + ….. see the attachment for further...
Nov 8, 2006 #7 T tomshack Junior Member level 3 Joined Mar 4, 2006 Messages 27 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,286 Activity points 1,474 pmonon said: if b is the indep. variable and w is const., 1/(1+(wb)2) = 1/(1+(wδ)2) - {2wδ/(1+(wδ)2 )-2}(b-δ) - ........ + ….. see the attachment for further... Click to expand... thank u my friend but i have the result but i dont know the solution .........the question is 9.th in pdf file
pmonon said: if b is the indep. variable and w is const., 1/(1+(wb)2) = 1/(1+(wδ)2) - {2wδ/(1+(wδ)2 )-2}(b-δ) - ........ + ….. see the attachment for further... Click to expand... thank u my friend but i have the result but i dont know the solution .........the question is 9.th in pdf file
Nov 9, 2006 #8 T tomshack Junior Member level 3 Joined Mar 4, 2006 Messages 27 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,286 Activity points 1,474 ok guys i solved my problem thnk u for all
Nov 10, 2006 #9 E-design Advanced Member level 5 Joined Jun 1, 2002 Messages 2,014 Helped 507 Reputation 1,016 Reaction score 408 Trophy points 1,363 Activity points 14,193 Care to let us in on the solution?
Nov 10, 2006 #10 T tomshack Junior Member level 3 Joined Mar 4, 2006 Messages 27 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,286 Activity points 1,474 E-design said: Care to let us in on the solution? Click to expand... sure the taylor expansion of 1/1+x² = Σ ((-1)^n ) (x^2*n) n=0:finite wb=a in the question ....sorry e desing i didnt tell u wb is a constant at the first post...thank u for all your helps
E-design said: Care to let us in on the solution? Click to expand... sure the taylor expansion of 1/1+x² = Σ ((-1)^n ) (x^2*n) n=0:finite wb=a in the question ....sorry e desing i didnt tell u wb is a constant at the first post...thank u for all your helps