Find the coordinate matrix of the function $\cos^2x$ relative to the ordered basis $\{1,\cos x,\sin x,\sin 2x\}$.
Find the coordinate matrix of the function $\cos^2x$ relative to the ordered basis $\{1,\cos x,\sin x,\sin 2x\}$.
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linear-algebra
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2This sounds like a command, not a question. You should give us some context for your problem, as well as what your ideas are so far. – 2012-12-15
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1Since when are sets ordered? – 2012-12-15
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0@akkkk: Sets are not generally ordered, but a basis is ordered so that the co-ordinates read off the basis becomes unique. – 2012-12-15
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1Since $\cos^2(x)$ is even and has period $\pi$ it is not an exact linear combination of the basis elements. If this is about approximation then this question needs more context. – 2012-12-15
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0I see that {1,sinx,cosx} generates cos^2 but how ı can find coordinate matrix? – 2012-12-15
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0@signum Then you see more than I do. – 2012-12-15
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0I don't think that given functions form a basis, since last element depends on third and second as $\sin 2x = 2 \sin x \cos x$ – 2012-12-15
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0@Kaster How's that a linear dependency? – 2012-12-15
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0@WimC I don't understand your question. – 2012-12-15
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0OK, gotcha, those are linearly independent. My bad. – 2012-12-15
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0Should that $\sin 2x$ maybe be $\sin^2 x$? – 2012-12-15
1 Answers
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Assume $$\tag1\cos^2x=a+b\cos x+c\sin x+d\sin 2x.$$ Then $$\tag 2\cos 2x = 2\cos^2x-1=(2a-1)+2b\cos x+2c\sin x+2d\sin 2x.$$ Taking derivatives of $(2)$, we obtain $$\tag3-2\sin 2x =-2b\sin x+2c\cos x+4d\cos 2x.$$ Solving $(3)$ for $2d\cos2x$, we find $$\tag42d\cos 2x=b\sin x-c\cos x-\sin 2x$$ and by equating $(4)$ and $2d(2)$ $$\tag5 0=2d(2a-1)+(4d-1)b\cos x+(4d+1)c\sin x+(4d^2+1)\sin2x.$$
Taking the linear independece of $\{1, \cos x,\sin x,\sin 2x\}$ for granted (which is correct), we read $4d^2+1=0$ from $(5)$, which is not possible in the reals. But even allowing complex coefficients, we also conclude $b=c=0$ and $a=\frac12$, which contradicts what we obtain from letting $x=0$ in $(1)$, namely $1=a+b$.