Proof: Smallest Positive Number - Inequality.

Question

Let $x_0$ be the smallest positive number such that $u(x_0)=0$ where $u$ is the solution for
$$u=\sum_{k=0}^{\infty }a_kx^k$$$$u''+x^2u=0$$$$u(0)=1, u'(0)=0$$ (I have a solution for the power series). $$u''+x^2 u=0\implies \sum_{n=0}^\infty n(n-1)a_nx^{n-2}+\sum_{n=0}^\infty a_nx^{n+2}=0$$ Prove that $$1-\frac{x^4}{12}\leq u(x)\leq 1$$ for $$0\leq x\leq x_0$$$$1-\frac{x^2}{12}\leq u(x)\leq 1$$$$x\epsilon (0,x_0)$$$$u''=-x^2u$$$$u(0)=-1$$$$u(x)> 0$$$$x\rightarrow x_0$$$$u''=-x^2u< 0$$ $$x\rightarrow 0 , (0,x_0)$$$$u'\rightarrow 0 ,(0,x_0)$$$$u'(0)=0$$$$u'(x)\leq u'(0)=0 ,(0,x_0)$$ $u$ decrease in $(0,x_0)$ $$u''=-x^2u, (0,x_0)$$$$u=1, u''=-x^2$$$$\int \int 1-\frac{x^2}{12}dx=$$ But where to from here?

Answer 1

You get a power series $$f(z)=\sum_k (-1)^k c_kz^k $$ (here $z=x^4$, $u(x)=f(x^4)$) with $c_0=1$, $c_k=q_1q_2·…·q_k$, $1>q_{j+1}>q_j\to0$ for $j\to\infty$.

This means that for any $N$ and $R_{N+1}=q_N^{-1}$ the partial sequence of $c_kz^k$ with $k>N$ is monotonically decreasing for $0\le z<R_{N+1}$. By the Leibniz test and its error bounds, the power series converges for $0\le z<R_{N+1}$ and the value of the power series is contained between the partial sums $s_{N-1}(z)$ and $s_{N}(z)$ as \begin{align} (-1)^N\left(f(z)-s_{N-1}(z)\right) &=\sum_{k=0}^\infty (-1)^kc_{N+k}z^{N+k} \\ &=\sum_{k=0}^\infty c_{N+2j}z^{N+2j}\left(1-q_{N+2j+1}z\right) &&(\ >0\ \ )\\ &=c_{N}z^N-\sum_{k=0}^\infty c_{N+2j+1}z^{N+2j+1}\left(1-q_{N+2j+2}z\right)&&(\ <c_{N}z^N\ ). \end{align}

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In my answer https://math.stackexchange.com/a/2348258/115115 to your previous question there are more details on this problem. For $q_k=\frac1{4k(4k-1)}$, $R=\sqrt7$ (or $N=1$, $|x|<\sqrt[4\,]{R_2}=\sqrt[4\,]{56}$) you get $$ 1-\frac{x^4}{3·4}=s_1(x^4)< u(x)<s_2(x^4)= 1-\frac{x^4}{3·4}\left(1-\frac{x^4}{7·8}\right)<s_0(x^4)=1 $$ and the root is contained between the roots of the polynomials on both sides, $$ 2·\sqrt[4\,]{1-\frac14}<x_0<2·\sqrt[4\,]{1+\frac1{2(3+\sqrt7)}}. $$