Wai Posted March 9, 2008 Posted March 9, 2008 How do I express 1/(x^4 + 1) as a partial fraction, without using imaginary numbers? I'm having trouble finding the values of a and b.
thedarkshade Posted March 9, 2008 Posted March 9, 2008 [math]x^4 +1[/math] is the same as [math]({x^2})^2 + 1^2[/math] and a more general formula would be [math]a^2 + b^2=(a \pm b)^2 \mp 2ab[/math]!
Country Boy Posted March 11, 2008 Posted March 11, 2008 How do I express 1/(x^4 + 1) as a partial fraction, without using imaginary numbers? I'm having trouble finding the values of a and b. What do you mean by "using" imaginary numbers? If you just mean that you don't want imaginary numbers in the final result, Here's how I would do it: the equation [math]x^4+ 1= 0[/math] has roots [math]\frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2}[/math], [math]\frac{\sqrt{2}}{2}- i\frac{\sqrt{2}}{2}[/math], [math]-\frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2}[/math], and [math]-\frac{\sqrt{2}}{2}- i\frac{\sqrt{2}}{2}[/math] so [math]x^4+ 1= (x-(\frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2}))(x- \frac{\sqrt{2}}{2}- i\frac{\sqrt{2}}2})}(x- (-\frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2}x-(\frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2}[/math] Now write those factors as [math](x-\frac{\sqrt{2}}{2}- i\frac{\sqrt{2}}{2})(x-\frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2}= (x-\frac{\sqrt{2}}{2})^2+ \frac{1}{2}[/math] and [math](x+\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})(x+ \frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2})= (x+ \frac{\sqrt{2}}{2})^2+ \frac{1}{2}[/math] which have only real coefficients.
Wai Posted March 11, 2008 Author Posted March 11, 2008 Thanks! I think I've got the right answer now.
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