Home page for accesible maths 1 1 Further Integration 1.9 Form of partial fractions 1.11 Continuation of proof

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1.10 Sketch of proof of Theorem 1.9 (non-examinable)

Proof. As in 1.6, we factorize $g(x)$ :

g(x)=(x-a_{1})^{m_{1}}\ldots(x-a_{s})^{m_{s}}Q_{1}(x)^{n_{1}}\ldots Q_{t}(x)^{% n_{t}}

where $a_{1},\ldots,a_{s}$ are distinct real numbers and $Q_{1}(x),\ldots,Q_{t}(x)$ are distinct irreducible quadratic polynomials. We will prove the theorem by induction on $s+t$ .

Assuming $s\geq 1$ then we consider the polynomials $(x-a_{s})^{m_{s}}$ and

g_{1}(x)=(x-a_{1})^{m_{1}}\ldots(x-a_{s-1})^{m_{s-1}}Q_{1}(x)^{n_{1}}\ldots Q_% {t}(x)^{n_{t}}.

Then $(x-a_{s})^{m_{s}}g_{1}(x)=g(x)$ . If $t\geq 1$ then we set

g_{2}(x)=(x-a_{1})^{m_{1}}\ldots(x-a_{s})^{m_{s}}Q_{1}(x)^{n_{1}}\ldots Q_{t-1% }(x)^{n_{t-1}}

so that $g_{2}(x)Q_{t}(x)^{n_{t}}=g(x)$ .