# am trying to calculate the global error bound for Euler's method, but I am having trouble. I am given the formula |y(t_i)−u_i|<=1L(hM/2+delta/h)(e^(L(t_i−a))−1)+|delta_0|e^(L(t_i−a)) where ui is the Euler approxmation. I am also given M,L,a,delta,delta_0, h. If I am not mistaken this will give me the error for each step, but how do I find the upper bound for the total error?

I am trying to calculate the global error bound for Euler's method, but I am having trouble. I am given the formula $|y\left({t}_{i}\right)-{u}_{i}|\le \frac{1}{L}\left(\frac{hM}{2}+\frac{\delta }{h}\right)\left({e}^{L\left({t}_{i}-a\right)}-1\right)+|{\delta }_{0}|{e}^{L\left({t}_{i}-a\right)}$ where ${u}_{i}$ is the Euler approxmation. I am also given $M,L,a,\delta ,{\delta }_{0}$, h. If I am not mistaken this will give me the error for each step, but how do I find the upper bound for the total error?
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Rebecca Benitez
${\mathrm{sin}}^{-1}\left(\frac{2x}{1+{x}^{2}}\right)=\theta \phantom{\rule{1em}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{1em}{0ex}}\mathrm{sin}\left(\theta \right)=\frac{2x}{1+{x}^{2}}\phantom{\rule{1em}{0ex}}\wedge \phantom{\rule{1em}{0ex}}-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$
Note that
$\mathrm{cos}\left(\theta \right)=|\frac{1-{x}^{2}}{1+{x}^{2}}|$
When $|x|\le 1$
$\mathrm{tan}\left(\theta /2\right)=\frac{\mathrm{sin}\left(\theta \right)}{1+\mathrm{cos}\left(\theta \right)}=\frac{\frac{2x}{1+{x}^{2}}}{1+\frac{1-{x}^{2}}{1+{x}^{2}}}=x$
Therefore,
$\theta =2{\mathrm{tan}}^{-1}\left(x\right)$
When $|x|>1$
$\mathrm{tan}\left(\theta /2\right)=\frac{\mathrm{sin}\left(\theta \right)}{1+\mathrm{cos}\left(\theta \right)}=\frac{\frac{2x}{1+{x}^{2}}}{1-\frac{1-{x}^{2}}{1+{x}^{2}}}=1/x$
Therefore,
$\theta =2{\mathrm{tan}}^{-1}\left(1/x\right)$
Thus,

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