quick q - binomial or taylor (1+x)^-1 types... Watch

xfootiecrazeesarax
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For something like

 \frac{1}{1+3x}

you can get it into the form  1/1+x = 1 -x + x^{2} + ... *

in order to use that expansion

i.e.  \frac{1}{1+3x} =  \frac{1}{3(1+\frac{x}{3})}


Instead if you have something like  \frac{1}{3+x} , is there something similar that you could do in order to use * or do you need to use taylor expansion instead?

Many thanks
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RDKGames
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(Original post by xfootiecrazeesarax)
For something like

 \frac{1}{1+3x}

you can get it into the form  1/1+x = 1 -x + x^{2} + ... *

in order to use that expansion

i.e.  \frac{1}{1+3x} =  \frac{1}{3(1+\frac{x}{3})}


Instead if you have something like  \frac{1}{3+x} , is there something similar that you could do in order to use * or do you need to use taylor expansion instead?

Many thanks
\frac{1}{1+3x} \not=  \frac{1}{3(1+\frac{x}{3})} - here you simply do x\mapsto 3x on the Taylor expansion

and for \frac{1}{3+x} you would turn it into \frac{1}{3}\cdot \frac{1}{1+\frac{x}{3}} and do x\mapsto \frac{x}{3}
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