# New Methods of Series Expansions between Three Anomalies
## 2 Materials and Methods
### The Geometric Relationship among True Anomaly, Elliptic Anomaly, and Mean Anomaly
If there are two variables with the following relationship,  
\[\tan y=K\tan x \tag{9}\]  
Then, there can be the following relationship:  
\[y-x=\frac{K-1}{K+1}\sin 2y-\frac{1}{2}\bigg{(}\frac{K-1}{K+1}\bigg{)}^{2}\sin 4y +\frac{1}{3}\bigg{(}\frac{K-1}{K+1}\bigg{)}^{3}\sin 6y\cdots \tag{10}\]  
Alternatively, there can also be the following forms:  
\[y-x=\frac{K-1}{K+1}\sin 2x+\frac{1}{2}\bigg{(}\frac{K-1}{K+1}\bigg{)}^{2}\sin 4 x+\frac{1}{3}\bigg{(}\frac{K-1}{K+1}\bigg{)}^{3}\sin 6x\cdots \tag{11}\]  
According to Formula (7) of the relationship between the true anomaly and the elliptic anomaly and the formula of the Lagrange conjugate series, let \(\sqrt{\frac{1+e}{1-e}}=K\), \(\frac{K-1}{K+1}=m\), and then we can obtain the following two Formulas (12) and (13):  
\[\frac{f}{2}-\frac{E}{2}=m\sin f-\frac{1}{2}m^{2}\sin 2f+\frac{1}{3}m^{3}\sin 3f\cdots \tag{12}\]  
\[\frac{f}{2}-\frac{E}{2}=m\sin E+\frac{1}{2}m^{2}\sin 2E+\frac{1}{3}m^{3}\sin 3E\cdots \tag{13}\]  
We can further derive the following formulas by simply deriving the two Formulas (12) and (13) above:  
\[E=f-2\sum_{p=1}^{\infty}(-1)^{p+1}\frac{1}{p}m^{p}\sin pf \tag{14}\]  
\[f=E+2\sum_{p=1}^{\infty}\frac{1}{p}m^{p}\sin pF \tag{15}\]In Formulas (14) and (15),  
\[=\sqrt{\frac{1+e}{1-e}}\tan\frac{E}{2}\]0  
By substituting Equations (16) into Equations (14) and (15), respectively, we can obtain the trigonometric series expansion formula with eccentricity \(e\) as the parameter. In the case where \(e\) is equal to zero, we expand the elliptic anomaly \(E\) and the true anomaly \(f\) into a power series form of eccentricity \(e\). We take the expanded formulas to the \(e^{8}\) terms to obtain Equations (17) and (18),  
\[=\sqrt{\frac{1+e}{1-e}}\tan\frac{E}{2}\]1  
\[=\sqrt{\frac{1+e}{1-e}}\tan\frac{E}{2}\]2  
The coefficients in Formulas (17) and (18) are as follows:  
\[\left\{ \begin{array}{l}a_{1}=-e-\frac{e^{3}}{4}-\frac{e^{5}}{8}-\frac{e^{5}}{ 64}\\ a_{2}=\frac{e^{2}}{4}+\frac{e^{4}}{4}+\frac{5e^{6}}{64}+\frac{7e^{8}}{128}\\ a_{3}=-\frac{e^{3}}{12}-\frac{e^{3}}{16}-\frac{e^{3}}{84}\\ a_{4}=\frac{e^{2}}{32}+\frac{e^{2}}{256}\\ a_{5}=-\frac{e^{5}}{80}-\frac{e^{5}}{84}\\ a_{6}=\frac{e^{8}}{192}+\frac{e^{3}}{128}\\ a_{7}=-\frac{e^{7}}{448}\\ a_{8}=\frac{e^{8}}{1024}\end{array} \right. \tag{19}\]  
\[\left\{ \begin{array}{l}b_{1}=e+\frac{e^{3}}{4}+\frac{e^{5}}{8}+\frac{5e^{7}} {64}\\ b_{2}=\frac{e^{2}}{4}+\frac{e^{4}}{8}+\frac{5e^{6}}{64}+\frac{7e^{8}}{128}\\ b_{3}=\frac{e^{2}}{12}+\frac{e^{2}}{16}+\frac{3e^{4}}{64}\\ b_{4}=\frac{e^{2}}{32}+\frac{e^{2}}{32}+\frac{e^{2}}{256}\\ b_{5}=\frac{e^{5}}{80}+\frac{e^{5}}{64}\\ b_{6}=\frac{e^{6}}{1024}+\frac{e^{8}}{128}\\ b_{7}=\frac{e^{7}}{448}\\ b_{8}=\frac{e^{7}}{1024}\end{array} \right. \tag{20}\]  
The Derivation of the Expansion Formula for the Relationship between Elliptic Anomaly and Mean Anomaly  
By shifting the terms of [PERSON]'s Equation (8), we can obtain the following symbolic iterative relationship (14) [35]:  
\[E=M+e\sin E \tag{21}\]  
Due to the small difference between the elliptic anomaly \(E\) and the mean anomaly \(M\), the initial value \(E_{0}\) of the elliptic anomaly \(E\) can be taken as \(E_{0}=M\). By substituting the initial value \(E_{0}\) of the elliptic anomaly into Formula (21), we can obtain the first iteration value \(E_{1}\) of the elliptic anomaly:  
\[E_{1}=M+e\sin E_{0} \tag{22}\]  
By substituting the first iteration value \(E_{1}\) of the elliptic anomaly into [PERSON]'s Equation (21) for the second iteration, we expand the formula to the 8 th order of \(e\) (If you want to improve computational accuracy, you need to expand to a higher order of \(e\)). We can obtain the second iteration result of the elliptic anomaly as follows:  
\[E_{2}=M+e\sin E_{1} \tag{23}\]The second iteration value \(E_{2}\) of the elliptic anomaly is substituted into [PERSON]'s Equation (21) to continue the iteration. Through calculation, we found that when the iteration formula for the elliptic anomaly is iterated for the ninth time, the coefficients of the formula no longer change.