Question

Use Newton's method to find the coordinates of the inflection point of the curve?

`y=e^cosx` 0 ≤ x ≤ π , correct to six decimal places.

Answer 1

`x=0.904557` to 6dp corresponds to an inflection point.

Explanation:

First let us look at the graph `y = f(x) = e^(cosx)`

Inflection points are the points of the curve where the curvature changes its sign, and from the graph I would predict there is one point `0 lt x lt pi/2`.

A necessary condition for an inflection point is that `f''(x)=0` and a sufficient existence condition requires `f′′(x + epsilon)` and `f′′(x - epsilon)` to have opposite signs in the `epsilon`-neighbourhood of `x`.

So if:

`f(x) = e^(cosx)`

Then using the chain rule we have

`f'(x) = e^(cosx)*(-sinx)`
`" "= -sinx e^(cosx)`

And as we need the second derivative we differentiate again (using the product rule) to get:

`f''(x) = (-sinx)(-sinx e^(cosx)) + (-cosx)(e^(cosx))`
`" "= sin^2x \ e^(cosx) -cosx \ e^(cosx)`
`" "= (sin^2x -cosx ) e^(cosx)`

Our necessary condition is `f''(x) = 0`, so this requires that:

`(sin^2x -cos x) e^(cosx) =0`

And as `e^x >0 AA x in RR`; then we must have:

`sin^2x -cos x =0`

So our aim is to solve the equation `g(x)=f''(x)=0` We can find the solution numerically, and using the Newton-Rhapson method we use the following iterative sequence

`{ (x_0,=1), ( x_(n+1), = x_n - g(x_n)/(g'(x_n)) ) :}`

We therefore need `g'(x)`, ie `f'''(x)`. so differentiating the above result a further time we get:

`\ \ g(x) = (sin^2x -cosx ) e^(cosx)`

`g'(x)= (sin^2x -cosx)(-sinx e^(cosx)) + (2sinxcosx-sinx)(e^(cosx))`
`" " = -sinx \ e^(cosx)( sin^2x -cosx -2cosx-1)`
`" " = -sinx \ e^(cosx)( sin^2x -3cosx-1)`

Then using excel working to 8dp we can tabulate the iterations as follows:

We have confirmed our necessary condition for an inflection point but we need to establish the necessary condition: that is `f′′(x + epsilon)` and `f′′(x - epsilon)` to have opposite signs in the `epsilon`-neighbourhood of `x`.

And using a calculator we can easily verify that with `epsilon=0.1` we have:

`f''(x)~~0`, `f''(x-epsilon) <0` and `f''(x+epsilon) >0`
And we conclude that the solution is `x=0.904557` to 6dp

Hence `x=0.904557` to 6dp corresponds to an inflection point, which is consistent with our initial prediction.

Exact Solution
NB: We can solve our equation for an exact answer

`(1-cos^2x) -cos x =0`
`:. cos^2x +cos x -1 =0`

This is a quadratic in `cos x`, so we can complete the square to get:

`(cos x+1/2)^2 -(1/2)^2-1 =0`
`:. (cos x+1/2)^2 =5/4`
`:. cos x+1/2 =+-1/2sqrt(5)`
`:. cos x =1/2(sqrt(5)-1) " " (because 0 le x le pi)`
`:. cos x =0.61803...`
`:. x =0.90455...`

Bear this in mind when problem solving as N-R was actually harder than solving the equation directly