J_1(x) = (x^2/10)*(J_1(x) + J_3(x)) How to solve?

[qnach]

Homework Statement

This is not a homewrok

Relevant Equations

J[1, x] = (x^2/10)*(J[1, x] + J[3, x])

Has any one any idea to solve this equation J₁(x) = (x²/10)*(J₁(x) + J₃(x)), in which J are spherical Bessel function normally write as $j_1 (x)$ and $j_3(x)$

Methods 1 serial expansion:
$j_1(x) = \frac{\sin(x)}{(x)^2} - \frac{\cos(x)}{x} \approx \dfrac{x}{3} - \dfrac{(x)^3}{30} + \dfrac{(x)^5}{840} + ...$
$j_3 (x) \approx \dfrac{(x^3)}{105}$
I have
$\left(\frac{x}{3}-\frac{x^3}{30} + \frac{x^5}{840} \right) \approx \frac{x^2}{10} \left[ \frac{x}{3} - \frac{x^3}{30} + \frac{x^3}{105} \right]$
This will lead to
$x = \sqrt{ \frac{1 - \sqrt{1 - Y}}{Z} } \approx 0.79$
Y and Z are some complicate expression.

Methods using numerical method.
The result obtained is 2.27 or so....
They are different. Has anyone any idea about what is wrong....

 

[anuttarasammyak]

You forgot to put double dollar signs at head and tail of the equalion lines. I put them.
------------
Has any one any idea to solve this equation $J[1, x] = (x^2/10)*(J[1, x] + J[3, x])$, in which J are spherical Bessel function normally write as $j_1 (x)$ and $j_3(x)$

Methods 1 serial expansion:
$$j_1(x) = \frac{\sin(x)}{(x)^2} - \frac{\cos(x)}{x} \approx \dfrac{x}{3} - \dfrac{(x)^3}{30} + \dfrac{(x)^5}{840} + ...$$
$$j_3 (x) \approx \dfrac{(x^3)}{105}$$
I have
$$\left(\frac{x}{3}-\frac{x^3}{30} + \frac{x^5}{840} \right) \approx \frac{x^2}{10} \left[ \frac{x}{3} - \frac{x^3}{30} + \frac{x^3}{105} \right]$$
This will lead to
$$x = \sqrt{ \frac{1 - \sqrt{1 - Y}}{Z} } \approx 0.79$$
Y and Z are some complicate expression.

Methods using numerical method.
The result obtained is 2.27 or so....
They are different. Has anyone any idea about what is wrong....
------------

I don't think expansion is a good way. Why don't you express the equation directry by trigonometry fucntion and powers of x by using recurrence formula which reduces $j_3(x)$ to $j_0(x)$ and $j_1(x)$ which is made of trigonometry function and powers of x. I have got a rather simple equation.

[EDIT]

Homework Statement: This is not a homewrok
Relevant Equations: J[1, x] = (x^2/10)*(J[1, x] + J[3, x])

Has any one any idea to solve this equation J₁(x) = (x²/10)*(J₁(x) + J₃(x)), in which J are spherical Bessel function normally write as $j_1 (x)$ and $j_3(x)$

Methods 1 serial expansion:
$j_1(x) = \frac{\sin(x)}{(x)^2} - \frac{\cos(x)}{x} \approx \dfrac{x}{3} - \dfrac{(x)^3}{30} + \dfrac{(x)^5}{840} + ...$
$j_3 (x) \approx \dfrac{(x^3)}{105}$
I have
$\left(\frac{x}{3}-\frac{x^3}{30} + \frac{x^5}{840} \right) \approx \frac{x^2}{10} \left[ \frac{x}{3} - \frac{x^3}{30} + \frac{x^3}{105} \right]$
This will lead to
$x = \sqrt{ \frac{1 - \sqrt{1 - Y}}{Z} } \approx 0.79$
Y and Z are some complicate expression.

Double dollar, not a single dollar. Double sharp for ones in text lines.

 

[vela]

Methods using numerical method.
The result obtained is 2.27 or so....
They are different. Has anyone any idea about what is wrong....

Using Mathematica, I expanded $j_1(x) - \tfrac{x^2}{10}(j_1(x)+j_3(x))$ in a series and plotted the result keeping only the first three non-vanishing terms. This is what I got.

The function never crosses the axis for $x>0$, so I'm surprised you found $x \approx 0.79$. So one mistake seems to be lurking in how you found that value.

Here's a plot using the spherical Bessel functions (blue), the first four terms (orange), and the first five terms (green) of the series.

So your fundamental error apparently was not keeping enough terms in the series expansion. As you can see in the plots, the root is actually around 2.7. I assume when you wrote 2.27, it was a typo.

 

[Denisson]

The discrepancy arises because your series expansion (truncated at low order) is only valid for very small xx, but the actual root of the equation is much larger (x≈2.744x≈2.744). When you keep only the first few terms of the spherical Bessel series, you get a polynomial that has a small positive root (≈0.79≈0.79), but this is an artifact of the truncation—it does not approximate the true root of the original transcendental equation.

You can simplify the original equation using the recurrence relation for spherical Bessel functions:

jn−1(x)+jn+1(x)=2n+1x jn(x).jn−1(x)+jn+1(x)=x2n+1jn(x).
For n=2n=2:

j1(x)+j3(x)=5x j2(x).j1(x)+j3(x)=x5j2(x).
Substitute into your equation j1(x)=x210(j1(x)+j3(x))j1(x)=10x2(j1(x)+j3(x)):

j1(x)=x210⋅5x j2(x)=x2 j2(x).j1(x)=10x2⋅x5j2(x)=2xj2(x).
Thus the equation reduces to

2j1(x)=x j2(x).2j1(x)=xj2(x).
Using the explicit forms

j1(x)=sin⁡xx2−cos⁡xx,j2(x)=(3x3−1x)sin⁡x−3x2cos⁡x,j1(x)=x2sinx−xcosx,j2(x)=(x33−x1)sinx−x23cosx,
after simplification you obtain the transcendental equation

xcos⁡x+(x2−1)sin⁡x=0⟺tan⁡x=−xx2−1.xcosx+(x2−1)sinx=0⟺tanx=−x2−1x.
Solving this numerically (e.g., by root‑finding) gives the first positive non‑zero root at

x≈2.744.x≈2.744.
This is the correct solution to your original equation. Your numerical result of 2.272.27 is likely due to an error in the numerical implementation or misinterpretation of the functions. The series method fails because the root lies outside the radius where the low‑order truncation is meaningful. For accurate results, always use the exact recurrence or the simplified transcendental form.