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I have posted a link there to this thread so the OP can view my work.mangoqueen54 said:What are the intersections of y=sin(x) and y=1-x^2?

I have the Pi/4 side but i dont know what the negative one is

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I have posted a link there to this thread so the OP can view my work.mangoqueen54 said:What are the intersections of y=sin(x) and y=1-x^2?

I have the Pi/4 side but i dont know what the negative one is

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We are given the two curves:

\(\displaystyle y=\sin(x)\)

\(\displaystyle y=1-x^2\)

And we are asked to find the points of intersection.

If we equate the two curves, we get:

\(\displaystyle \sin(x)=1-x^2\)

Which we can arrange as:

\(\displaystyle \sin(x)+x^2-1=0\)

So, if we define:

\(\displaystyle f(x)=\sin(x)+x^2-1\)

We may then find its roots. We will need to use a numeric root finding method since we cannot explicitly solve for $x$. So, we will use Newton's method:

\(\displaystyle x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f'\left(n_{n}\right)}\)

Using our definition of $f$, we obtain:

\(\displaystyle x_{n+1}=x_{n}-\frac{\sin\left(x_{n}\right)+x_{n}^2-1}{\cos\left(x_{n}\right)+2x_{n}}=\frac{x_{n}\cos\left(x_{n}\right)-\sin\left(x_{n}\right)+x_{n}^2+1}{\cos\left(x_{n}\right)+2x_{n}}\)

Next, let's look at a plot of our function $f$ and see how many roots there are and get a rough estimate of their values:

Now, ignoring for the moment the root approximations shown on the graph, let's just say we see that the smaller root is about -1.5 and then use Newton's recursive method to get an accurate approximate of this smaller root

\(\displaystyle x_0=-1.5\)

\(\displaystyle x_1\approx-1.41379912599863\)

\(\displaystyle x_2\approx-1.40963375165233\)

\(\displaystyle x_3\approx-1.40962400405597\)

\(\displaystyle x_4\approx-1.40962400400260\)

\(\displaystyle x_5\approx-1.40962400400260\)

Our last two successive approximations agree to 15 digits, so let's now find the other root, which we see is near x=0.5:

\(\displaystyle x_0=0.5\)

\(\displaystyle x_1\approx0.644107890053782\)

\(\displaystyle x_2\approx0.636750907010919\)

\(\displaystyle x_3\approx0.636732650918014\)

\(\displaystyle x_4\approx0.636732650805282\)

\(\displaystyle x_5\approx0.636732650805282\)

Our last two successive approximations agree to 15 digits, so we now have the two root approximations:

\(\displaystyle x\approx-1.40962400400260,\,0.636732650805282\)

Let's now verify that they are close approximations:

\(\displaystyle \sin(-1.40962400400260)\approx-0.987039832660\)

\(\displaystyle 1-(-1.40962400400260)^2\approx-0.987039832660\)

\(\displaystyle \sin(0.636732650805282)\approx0.594571531399\)

\(\displaystyle 1-(0.636732650805282)^2\approx0.594571531399\)

And so, using 12 decimal places of accuracy, we have found the points of intersection may be approximated by:

\(\displaystyle \bbox[10px,border:2px solid #207498]{(x,y)\approx(-1.409624004003,-0.987039832660),\,(0.636732650805,0.594571531399)}\)