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My question is:

1a) Show that the function $f(x)=x^3+\sin({\pi x \over 2})−1$,for $x \in [0,1]$,has a unique root $x^* \in [0,1]$.

How would I show this using numerical methods?

b) Write down the function g for the Newton–Raphson method to compute the root $x^*$, in the form:

$x_0 \in [0,1]$ and $x_k =g(x_{k-1})$ for k≥1

Is it reasonable to expect convergence?

I'm really struggling with how to answer these questions

Thanks in advance for any help!

copper.hat
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Mai Muop Nguyen
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1 Answers1

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Hint: The derivative is $f’(x)=3x^2+\frac{\pi}{2}\cos{\frac{\pi}{2}x}$.

  1. Justify that $f’(x)\ge 0$ over $[0,1]$. What can you deduce for $f$ (monotonicity)?

  2. what is the sign of $f(0)$ and $f(1)$. What can you deduce using Intermediate Value Theorem.

  3. Now use this lemma: A monotonic continuous function which changes sign over an interval $K$ vanishes a single time.

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DINEDINE
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  • Thank you for your help! I've calculated that f'0)=π/2 and f'(1)=3 which means that f'(x)≥0 over [0,1], therefore f is an increasing function. f(0)=-1 and f(1)=1 which implies f(0)<0<f(1) so it follows by the intermediate value theorem that f(x*)=0 for some x∗∈[0,1] -> f(x)=0 has a root between 0 and 1. Is this correct? – Mai Muop Nguyen Apr 07 '19 at 21:29
  • @MaiMuopNguyen $f’(0)=\pi/2$ and $f’(1)=3$ does not suffice to show that $f’(x)\ge 0$. Use this $\forall x\in [0,1], 0\le \frac{\pi}{2}x\le \frac{\pi}{2}$ therefore the cosine is positive. – DINEDINE Apr 07 '19 at 21:41
  • $x^2\ge 0$ and $\cos(\frac{\pi}{2}x)\ge 0$ thus $f’(x)\ge 0$. The rest is correct – DINEDINE Apr 07 '19 at 21:49