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

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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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    (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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