The error for approximation of bessel function 
The sixth degree polynomial
$1-{x^2\over 4}+{x^4\over 64}-{x^6\over 2304}$
Is sometimes used to approximate the Bessel function $J_0(x)$ of the first kind of order zero for $0 \leq x\leq 1$. Show that the error $E$ involved in this approximation is less than $0.00001$.

I know that
$J_0(x)=\sum_{n=0}^\infty \frac{(-1)^n}{(n!)^2} ({\frac{x}{2}})^{(2n)}= 1-{x^2\over 4}+{x^4\over 64}-{x^6\over 2304}+{x^8\over 147456}+...$
I think, We want to show that
$J_0(x)-( 1-{x^2\over 4}+{x^4\over 64}-{x^6\over 2304})\leq {x^8\over 147456}$
But How can I show that, and how can I use the information $0 \leq x\leq 1$, please?
“ this question in calculus course”
 A: The series is alternating and the ratio of the consecutive terms is
$$\frac {a_{n + 1}} {a_n} = -\frac {x^2} {4 (n + 1)^2}.$$
The absolute values of the terms converge to zero monotonically for any $x \in (0, 1]$, therefore
$$\left| J_0(x) - \sum_{k = 0}^n a_k \right| < |a_{n + 1}|.$$
Now observe that the denominator of $a_4$ is greater than $10^5$.
A: Statement:
$E(x)=J_0(x)-\sum\limits_{n=0}^3\frac{(-1)^n}{(n!)^2} (\frac{x}{2})^{2n}\leq\frac{x^8}{2^8 (4!)^2}\tag{1}$ 
$E(x)=\sum\limits_{n=4}^\infty\frac{(-1)^n}{(n!)^2} ({\frac{x}{2}})^{2n}=\sum\limits_{n=0}^\infty\frac{(-1)^{n+4}}{(n+4)!^2} ({\frac{x}{2}})^{2(n+4)}=\frac{x^8}{2^8}\sum\limits_{n=0}^\infty\frac{(i)^{2n}}{(n+4)!^2} ({\frac{x}{2}})^{2n}$ $\hspace{0,5cm}$ where $\hspace{0,2cm}$ $i^2=-1$
Enough to prove that $\hspace{0,2cm}$ $\sum\limits_{n=0}^\infty\frac{(i)^{2n}}{(n+4)!^2} ({\frac{x}{2}})^{2n} \le \frac{1}{(4!)^2}\tag{2}$
From other hand the following inequality is true for every $k\ge1 $ (integer) and inside the domain of $x$ is defined by $\hspace{0,2cm}$ $0\le x\le 1$: 
$\prod\limits_{k=1}^{n}\frac{(k+4)^2}{(i\frac{x}{2})^4}\gt1 \tag{3}$
$\prod\limits_{k=1}^{n}\frac{(k+4)^2}{(i\frac{x}{2})^4}=\frac{(n+4)!^2}{(4!)^2 (i\frac{x}{2})^4}\gt 1$
Realign the inequality:  $\hspace{0,2cm}$
$\frac{1}{(4!)^2 (i\frac{x}{2})^2}\gt\frac{(i\frac{x}{2})^2}{(n+4)!^2}\tag{4}$
Take the sum of both sides:
$\sum\limits_{n=0}^\infty\frac{1}{(4!)^2 (i\frac{x}{2})^2}\gt\sum\limits_{n=0}^\infty\frac{(i\frac{x}{2})^2}{(n+4)!^2}=\sum\limits_{n=0}^\infty\frac{i^{2n}}{(n+4)!^2}(\frac{x}{2})^{2n}$
RHS of the inequality is the same the sum then the sum in (2). LHS of the inequality is geometric series equal to:
$\sum\limits_{n=0}^\infty\frac{1}{(4!)^2 (i\frac{x}{2})^2}=\frac{1}{(4!)^2}\frac{1}{1+\frac{4}{x^2}}\lt \frac{1}{(4!)^2}$ for every $x$
So
$E(x)\lt \frac{x^8}{2^8 (4!)^2}$ regarding that $0\le x \le 1$ the maximum error value is at $x=1$ where $E(x)\lt 10^{-5}$
Similar method can be seen on the link 
Prove that ${e\over {\pi}}\lt{\sqrt3\over{2}}$ without using a calculator.
from user90369.
