B Trouble with polynomial long division

SamRoss

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Summary
Having trouble with a simple polynomial long division problem.
I'm reading a book where the author gives the long division solution of ##\frac 1 {1+y^2}## as ##1-y^2+y^4-y^6...##. I'm having trouble duplicating this result and even online calculators such as Symbolab are not helpful. Can anyone explain how to get it?
 

fresh_42

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It is difficult to display here and of course you can always multiply to check the result. I once explained it here:
Maybe this helps. It is the same as a normal division, step by step.
 

SamRoss

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It is difficult to display here and of course you can always multiply to check the result. I once explained it here:
Maybe this helps. It is the same as a normal division, step by step.
Thank you for your reply but I do not see how that link is related to my particular problem. Maybe it's because I'm not familiar with vocabulary like "splitting fields". Anyway, I have done basic polynomial long division before but never when the order of the dividend was less than the divisor. My thinking is to use negative exponents in the quotient but I don't think that would lead to the author's result.
 

fresh_42

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##1 : (1+y^2) = 1 \longrightarrow \Delta = -y^2##
##-y^2 : (1+y^2) = -y^2 \longrightarrow \Delta = y^4##
##y^4 : (1+y^2) = y^4 \longrightarrow \Delta = -y^6##
etc.

There is no restriction what has to be divided by what. In this case we start with ##1/1## and always divide by ##1##.
 

SamRoss

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##1 : (1+y^2) = 1 \longrightarrow \Delta = -y^2##
##-y^2 : (1+y^2) = -y^2 \longrightarrow \Delta = y^4##
##y^4 : (1+y^2) = y^4 \longrightarrow \Delta = -y^6##
etc.

There is no restriction what has to be divided by what. In this case we start with ##1/1## and always divide by ##1##.
What is meant by ##1 \longrightarrow \Delta ## and all the other ## \longrightarrow \Delta ## steps?
 

fresh_42

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What is meant by ##1 \longrightarrow \Delta ## and all the other ## \longrightarrow \Delta ## steps?
That's what is more elaborated in the first example I gave you. The deltas are the differences, because long division is a sequence of subtractions.
##A:(1+y^2) = A \longrightarrow \Delta = -Ay^2## reads:
First divide ##A## by ##1## which yields the ##A## right of the equation sign.
Next we multiply ##(1+y^2)## with this ##A## which yields ##A+Ay^2##.
Now this has to be subtracted from what we have on the far most left: ##A- (A+Ay^2)= -Ay^2 =\Delta## which I wrote as ##\Delta##. It is our next far most left entry:
##-Ay^2 : (1-y^2)= -Ay^2 \longrightarrow \Delta = -Ay^2-(-Ay^2 \cdot (1+y^2))= -Ay^2 -(- Ay^2) -(- Ay^4) =Ay^4##
etc.

Finally we gather all intermediate results: ##A : (1+y^2) = A -Ay^2 + Ay^4 \mp \ldots##
 

SamRoss

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Got it. Thanks! I was dividing by ##y^2## instead of ##1##. I feel silly.
 

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