# Computing inner products with linearity in the first argument.

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!

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.

• Thank you for the fast reply @JohnD as well...it is much simpler than I thought XD...thanks for the explanation! Dec 15, 2014 at 16:30
• @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? Dec 16, 2014 at 4:10

\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}