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Consider the vector space $..$ $(\mathbb{P},\langle \cdot,\cdot \rangle)$ where the inner product is given by:

$$$$ $\langle p(x),q(x) \rangle = \displaystyle\int_{-\infty}^{\infty} p(x)q(x)e^{-x^2} dx$

It can be shown that the moments in this inner product space are:

$$$$ $\langle 1,1 \rangle=\sqrt{\pi},$ $$$$ $\langle x,1 \rangle=0,$ $$$$ $\langle x^2,1 \rangle=\dfrac{\sqrt{\pi}}{2},$ $$$$ $\langle x^3,1 \rangle = 0,$ $$$$ $\langle x^4,1 \rangle = \dfrac{3\sqrt{\pi}}{4},$ $$$$ $\langle x^5,1 \rangle = 0,$ $$$$ $\langle x^6,1 \rangle = \dfrac{15\sqrt{\pi}}{8}.$

Use these moments and the "linear in the first argument" property of inner products to compute $..$ $\langle 4x^2+3x+9,1\rangle$ and $\langle 32x^5-64x^3+24x,1\rangle$

The linear in the first argument property is as follows:

For all scalars $:$ $\alpha,\beta$ and the vectors $\vec{x},\vec{y},\vec{z}$:

$$$$ $<\alpha \vec{x}+\beta \vec{y},\vec{z}> = \alpha<\vec{x},\vec{z}> + \beta<\vec{y},\vec{z}>.$

So I am a bit confused as to how we would change the above inner products that are needed to be computed to fit the moments. I understand that the second value of the inner product is already 1, but the first inner product value isn't a formula that already fit the moments. Would I first plug in the inner products in the integral then integrate and then answer based on the outcome (being one of the moments)?

Any help is much appreciated!

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2 Answers 2

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Here, your basis vectors are powers of $x$ in the polynomial ring, think of them as vectors and not as functions, this is an infinite dimension vector space with basis vectors $x^0,x^1,x^2,...$. we usually just use $1$ for $x^0$.

So your first, using linearity, expands to is just $4<x^2,1>+3<x,1>+9<1,1>=4\cdot \frac {\sqrt \pi} 2+3\cdot 0+9\cdot \sqrt \pi =11\sqrt \pi $.

The second is similar.

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  • $\begingroup$ Thank you for the fast reply @JohnD as well...it is much simpler than I thought XD...thanks for the explanation! $\endgroup$
    – Jacob
    Dec 15, 2014 at 16:30
  • $\begingroup$ @Jacob Just a friendly reminder in case you don’t know how the site works: upvoting any and all helpful answers is the best "thank you". Perhaps you didn't have enough rep? But maybe you do now? $\endgroup$
    – JohnD
    Dec 16, 2014 at 4:10
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\begin{align}\langle 4x^2+3x+9,1\rangle&=\langle 4x^2,1\rangle + \langle 3x,1\rangle + \langle 9,1\rangle\\ &=4\langle x^2,1\rangle + 3\langle x,1\rangle + 9\langle 1,1\rangle\\ &=4\cdot{\sqrt{\pi}\over 2}+3\cdot 0+9\cdot\sqrt{\pi}\\ &=11\sqrt{\pi} \end{align}

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