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Removed specific error in proof.

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Luc Bijl 2024-01-03 13:50:53 +01:00
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1 changed files with 12 additions and 3 deletions
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docs/en/mathematics/ordinary-differential-equations/laplace-transform.md
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@ -99,13 +99,22 @@ $$
For large enough $s$, the case $n=1$ follows by integration by parts For large enough $s$, the case $n=1$ follows by integration by parts
$$ $$
\begin{align*} \mathcal{L}[f'](s) &= \int_0^\infty e^{-st} f'(t)dt, \\ &= \Big[e^{-st} f(t) \Big]_0^\infty + s\int_0^\infty e^{-st}f(t), \\ &= sF(s) - f(0) \end{align*}, \begin{align*}
\mathcal{L}[f'](s) &= \int_0^\infty e^{-st} f'(t)dt, \\
&= \Big[e^{-st} f(t) \Big]_0^\infty + s\int_0^\infty e^{-st}f(t), \\
&= sF(s) - f(0),
\end{align*}
$$ $$
suppose $\mathcal{L}[f^{k}](s) = s^k F(s) - \sum_{r=0}^{k-1} s^r f^{(k-1-r)}(0)$ is true for $k \in \mathbb{N}$, then by assumption suppose $\mathcal{L}[f^{k}](s) = s^k F(s) - \sum_{r=0}^{k-1} s^r f^{(k-1-r)}(0)$ is true for $k \in \mathbb{N}$. Then by assumption
$$ $$
\begin{align*} \mathcal{L}[f^{k+1}](s) &= \int_0^\infty e^{-st} f^{(k+1)}(t)dt , \\ &= \Big[e^{-st} f^{(k+1)}(t) \Big]_0^\infty + s\int_0^\infty e^{-st}f^{(k)}(t), \\ &= s \mathcal{L}[f^{(k)}] - f^{(k)}(0), \\ &= s \Big(s^k F(s) - \sum_{r=0}^{k-1} s^r f^{(k-1-r)}(0)\Big) - f^{(k)}(0), \\ &= s^{k+1} F(s) - \sum_{r=0}^{k} s^r f^{(k-r)}(0) \end{align*}. \begin{align*}
\mathcal{L}[f^{k+1}](s) &= \int_0^\infty e^{-st} f^{(k+1)}(t)dt, \\
&= \Big[e^{-st} f^{(k+1)}(t) \Big]_0^\infty + s\int_0^\infty e^{-st}f^{(k)}(t), \\
&= s \mathcal{L}[f^{(k)}] - f^{(k)}(0), \\ &= s \Big(s^k F(s) - \sum_{r=0}^{k-1} s^r f^{(k-1-r)}(0)\Big) - f^{(k)}(0), \\
&= s^{k+1} F(s) - \sum_{r=0}^{k} s^r f^{(k-r)}(0).
\end{align*}
$$ $$
## Examples ## Examples