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For being a guy who dreams of studying medicine one day, it hurts me to ask such a question because it makes me feel like a total idiot. Anyways... $$1 - \frac{1}{1+x^2} = \frac{x^2}{1+x^2}$$

I can go from the left part to the right part, but not the other way. I just don't understand what the trick is her. It's not obvious to me. I came a across this rewrite while doing integration by parts.

Any mental-math tip for me?

Srivatsan
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Algific
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3 Answers3

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It is a common trick in mathematics to add and subtract something, e.g. $$x\longrightarrow (x-y)+y.$$ The idea being that the quantity is the same, but it gives you more possible terms to manipulate and (hopefully) simplify. In particular this can help with fractions like yours, $$\frac{x}{x+y}\longrightarrow\frac{(x+y)-y}{x+y}=\frac{x+y}{x+y}-\frac{y}{x+y}=1-\frac{y}{x+y}.$$ While for fractions it is usually easy to see what $y$ is necessary to get something that is easily divisible by the denominator, in general the key is knowing how to pick the right $y$... but I don't think there's anything other than intuition (and practice) that can help with that.

Zev Chonoles
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4

Just add zero to the numerator:

$$\frac{x^2}{1+x^2} = \frac{(1 - 1) + x^2}{1+x^2} = \frac{1+x^2}{1+x^2}-\frac{1}{1+x^2} = 1 - \frac{1}{1+x^2}$$

Calle
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0

This is just polynomial division: if $A(x)$ and $B(x)$ are polynomials with $\deg(b) \ge 1$, then $A(x)/B(x)$ can always be expressed as $Q(x) + R(x)/B(x)$, with $\deg(R) < \deg(B)$.

Do you know how to do this?

TonyK
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