The linear backward Euler's Method is given by

${y}^{j+1}={y}^{j}+\tau {k}_{1}$

$(I-\tau J){k}_{1}=f({y}^{j}),J:={f}^{\prime}({y}^{j})$

To show, is that ${y}^{j+1}$ is equal to $\stackrel{~}{{y}^{j+1}}$ the Output of the Backward Euler's Method, if you proccess one Newton-Step with

$g-({y}^{j}+\tau f(g))=0$

for $\stackrel{~}{{y}^{j+1}}=g$

${g}^{0}={y}^{j}$

This is what I started with :

${g}^{1}={g}^{0}+\tau \frac{1}{I-\tau {f}^{\prime}({g}^{0})}f({g}^{0})$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{y}^{j+1}={y}^{j}+\tau \frac{1}{I-\tau {f}^{\prime}({y}^{j})}f({y}^{j})$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{y}^{j+1}-{y}^{j}=\tau \frac{1}{I-\tau {f}^{\prime}({y}^{j})}f({y}^{j})$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\tau {k}_{1}=\tau \frac{1}{I-\tau {f}^{\prime}({y}^{j})}f({y}^{j})$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{k}_{1}(I-\tau {f}^{\prime}({y}^{j}))=f({y}^{j})$

Did I miss something or is there anything wrong? Any help is very appreciated

${y}^{j+1}={y}^{j}+\tau {k}_{1}$

$(I-\tau J){k}_{1}=f({y}^{j}),J:={f}^{\prime}({y}^{j})$

To show, is that ${y}^{j+1}$ is equal to $\stackrel{~}{{y}^{j+1}}$ the Output of the Backward Euler's Method, if you proccess one Newton-Step with

$g-({y}^{j}+\tau f(g))=0$

for $\stackrel{~}{{y}^{j+1}}=g$

${g}^{0}={y}^{j}$

This is what I started with :

${g}^{1}={g}^{0}+\tau \frac{1}{I-\tau {f}^{\prime}({g}^{0})}f({g}^{0})$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{y}^{j+1}={y}^{j}+\tau \frac{1}{I-\tau {f}^{\prime}({y}^{j})}f({y}^{j})$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{y}^{j+1}-{y}^{j}=\tau \frac{1}{I-\tau {f}^{\prime}({y}^{j})}f({y}^{j})$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\tau {k}_{1}=\tau \frac{1}{I-\tau {f}^{\prime}({y}^{j})}f({y}^{j})$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{k}_{1}(I-\tau {f}^{\prime}({y}^{j}))=f({y}^{j})$

Did I miss something or is there anything wrong? Any help is very appreciated