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We have been told that $u'(t)= u(t)$ and $u(0)=1$ from this information we can conclude that $u(t) =e^x$

Show that $ u_k =(1+h)^k,k=0,1,...$ is an approximation for $u(kh) $ using Euler's method

My current thoughts are that h is the interval by which we increment, but I am not sure where to go from there.

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Call

$$ u_k = u(k h) ~~~~\mbox{for}~~~ k = 0, 1, \cdots $$

Euler's method can be expressed as

$$ u_{k + 1} = u_k + hu'_k = u_k + h (u_k) = (1 + h)u_k $$

You can easily figure out the solution to this expression

\begin{eqnarray} u_1 &=& (1 + h)u_0 \\ u_2 &=& (1 + h)u_1 = (1 + h)^2u_0 \\ u_3 &=& (1 + h)u_2 = (1 + h)^3u_0 \\ &\vdots& \\ u_k &=& (1 + h)u_{k - 1} = (1 + h)^k u_0 \end{eqnarray}

which reduces to $u_k = (1 + h)^k$ if $u_0 = u(t = 0) = 1$

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