# Help with Engineering Mathematics, Struggling to determine coefficients using methods detailed in this research paper

I am struggling with a curve fitting calculation below. the part I am stuck with is arguably the simplest part, which is determining the coefficients $$\\K_3$$ and $$\\K_4$$. It seems like it should be simple/straightforward but it's not clear cut at this stage. some clarity would be appreciated on what I am missing in terms of defining the coefficients of the equation assuming that $$\\p = 5$$ just to keep it simple. I am trying to calculate $$\\I_s$$ using equation (II.10) assuming that $$\\I_h$$ = 0 and $$\\I_s = I_0$$

if $$\lambda\ = \lambda\ _m$$ $$\sin(\omega t)$$

$$\lambda\ ^5\ = \lambda\ _m^{5}$$ $$\sin^{5}(\omega t)$$

using the trigonometric power formula: $$\lambda\ ^5\ = \lambda\ _m^{5}$$ $$\sin^{5}(\omega t)$$ = $$\dfrac{1}{16}(10\lambda_m^{5}sin^5(\omega t)-5\lambda_m^{5}sin(3wt)+\lambda_m^{5}sin(5wt))$$ note: I used wolfram alpha to complete this formula and couldn't figure out how to use subscript, for this equation $$\lambda\ = \lambda\ _m$$

I could use some help with determining the coefficients $$\\K_3$$ and $$\\K_4$$ what am I missing ? • You can simplify the problem by noting that $\sin\omega t = \text{Im}[\exp(j\omega t)]$. To find $K_3$ and $K_4$ you need two equations that have to determined from one. You can do that by equating powers of $\lambda_m$ after computing the RMS value of $i_s$. – Biswajit Banerjee Sep 5 '19 at 22:06
• Thank you for this @Biswajit, so in the case of the function $\lambda\ ^5\ = \lambda\ _m^{5}$ $\sin^{5}(\omega t)$ would be simplified to $\frac{j}{32}\lambda _m^{5}\ (exp(jwt) - exp(jwt))^{5}$ and to be clear, finding $\ K_4$ and $\ K_5$ would need a simultaneous equation/system of equations and that could be created by calculating the RMS $\ i_s$ using the expansion for $\lambda$ that I've just done? and use Euler's representation to simplify the calculations ? – Bradley D Sep 6 '19 at 15:03
• correction, the above expression should be $\frac{j}{32}\lambda _m^{5}\ (exp(jwt) - exp(-jwt))^{5}$ – Bradley D Sep 6 '19 at 15:35
• If $\lambda = \lambda_m \exp(j\omega t)$ then $\lambda^5 = \lambda_m^5 \exp(5j\omega t)$. You can extract the imaginary part after all the calculations have been completed with the exponential form. – Biswajit Banerjee Sep 6 '19 at 22:01
• Silly me, we're saying similar things, so effectively what I need to do is find the RMS of the following expression $\ i_s = A_1\lambda + A_5\lambda^{5} = A_1(\lambda_m\exp(j\omega t))+A_5(\lambda_m^{5}exp(5j\omega t))$ and then solve the simultaneous equation of $\ i_{s(RMS)}$ and $\ I_s$ ? – Bradley D Sep 7 '19 at 1:42