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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”

Community
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Dima
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    You have a convergent alternating series with each term smaller than the previous one in absolute value. When you truncate it at even or odd terms you either overestimate or underestimate the value of the function. I think you are supposed to use this fact to estimate the error – Yuriy S Oct 26 '18 at 09:22
  • @YuriyS Thank you. – Dima Oct 27 '18 at 03:02

2 Answers2

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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 $10^5$.

Maxim
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2

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.

JV.Stalker
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