Let y = 1− t2 + 2t
The graph of `y(t)`:
There appear to be 2 roots, one near t = -1 and the other near t = 3. However, if we look more carefully in the region near t = 3 (by zooming in), we see that there is more than one root there.
By simple substitution, we can see that one of the roots is exactly t = 3:
1 − (3)2 + 23 = 0
Now for the root near t = 3.4.
We will use Newton's Method to approximate the root. We need to differentiate y = 1− t2 + 2t. Since we have t as an exponent in our expression, we need to use logarithms when differentiating. [See how to differentiate logarithms in Derivative of the Logarithmic Function].
Let's differentiate 2t by itself first.
Let h = 2t.
Take natural log of both sides:
`ln\ h=t\ ln\ 2`
`1/h (dh)/dt=ln\ 2`
`(dh)/(dt)=h\ ln\ 2=2^t\ ln\ 2`
`(dy)/(dt)=f'(t)=-2t+2^t\ ln\ 2`
So for Newton's Method in this example, we would have:
`(f(t))/(f'(t))=(1-t^2+2^t)/(-2t+2^t ln\ 2)`
Initial guess: t0 = 3.4
`t_1=3.4-(f(t))/(f'(t)) -` ` (1-3.4^2+2^3.4)/(-2(3.4)+2^3.4 ln\ 2) ` `= 3.407615918`
We can write this more conveniently (for later steps) showing the substitution as:
`t_1=3.4-[(1-t^2+2^t)/(-2t+2^t ln\ 2)]_(t=3.4)` `=3.407615918`
Now, doing another step of Newton's Method:
`t_2` `=3.407615918-` `[(1-t^2+2^t)/(-2t+2^t ln\ 2)]_(t=3.407615918)` `=3.407450596`
And doing another step:
`t_3` `=3.407450596-` `[(1-t^2+2^t)/(-2t+2^t ln\ 2)]_(t=3.407450596)` `=3.407450522`
`t_4` `=3.407450522-` `[(1-t^2+2^t)/(-2t+2^t ln\ 2)]_(t=3.407450522)` `=3.407450505`
We can conclude that correct to 7 decimal places, t = 3.4074505.Using Graphs Instead
Using Scientific Notebook, we can zoom into the root and we can see (from where the graph cuts the y-axis) that t is indeed close to `3.40745`.
Now for the negative case. Let t0 = −1 be our initial guess.
`t_1` `=-1-[(1-t^2+2^t)/(-2t+2^t ln\ 2)]_(t=-1)` `=-1.213076633`
`t_2` `=-1.213076633-` `[(1-t^2+2^t)/(-2t+2^t ln\ 2)]_(t=-1.213076633)` `=-1.198322474`
`t_3` `=-1.198322474-` `[(1-t^2+2^t)/(-2t+2^t ln\ 2)]_(t=-1.198322474)` `=-1.198250199`
We could continue until we obtained the required accuracy.
Comparing this to the zoomed in graph, we can see that the solution is t = -1.198250197, correct to 9 decimal places.
So the solutions for 1− t2 + 2t = 0 are
t = −1.19825,
t = 3, or
t = 3.40745,
correct to 5 decimal places.