Newton-Raphson - How many solutions?

Jan 2016
17
0
england
Hi, I was asked to find a solution to the following problem using the Newton-Raphson process:

x=(sin(x)+cos(x))^1/2

which I found to be 1.14955 (5dp)

But how to I determine how many solutions there are in total?

I plotted the graph and looked into tangents but think I'm barking up the wrong tree.. I can't quite get my head around this so any help would be greatly appreciated..! Thank you
 

v8archie

Math Team
Dec 2013
7,712
2,682
Colombia
Think about the range of possible values of $\sin{(x)} + \cos{(x)}$. This then gives you a range of $x$ within which all possible solutions must occur.

Note that $\sin{(x)} + \cos{(x)} = \sqrt{2}\left( \frac{1}{\sqrt2}\sin{(x)} + \frac{1}{\sqrt2}\cos{(x)} \right)$ can be expressed as a single sinusoidal function using $\sin{(A+B)} = \sin{(A)}\cos{(B)} + \cos{(A)}\sin{(B)}$
 
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Jan 2016
17
0
england
Thank you v8archie but I have I admit that I'm still very lost. I can see that the range is between –√2 and √2 but I can't see how you managed to manipulate that formula into that trig identity, nor do I understand why?

Sorry if i'm being stupid, I think I've just been staring at this problem for too many hours so nothing is making sense anymore.

Would you mind explaining a little bit more please?
 
May 2016
1,310
551
USA
Before we get to your question

$-\ 1 \le sin(x) \le 1\ and\ -\ 1 \le cos(x) \le 1 \implies$

$-\ 2 \le sin(x) + cos(x) \le 2 \implies$

$\sqrt{sin(x) + cos(x)} \in \mathbb R\ only\ if\ 0 \le sin(x) + cos(x) \le 2 \implies$

$0 \le x \le \sqrt{2}\ if\ x \in \mathbb R\ and\ x = \sqrt{sin(x) + cos(x)}.$

But your answer implies that $x > \sqrt{2}.$

So I am confused from the get-go.

Now consider the potential x-intercepts in $[0,\ \sqrt{2}]$ for the function

$f(x) = \sqrt{sin(x) + cos(x)} - x \implies f(0) = 1.$

Moreover, $f'(x) = -\ \dfrac{1}{2\sqrt{sin(x) + cos(x)}} - 1 < 0.$

So what is the maximum number of possible x-intercepts in $[0,\ \sqrt{2}]$
 
Last edited:

skeeter

Math Team
Jul 2011
3,363
1,854
Texas
x=(sin(x)+cos(x))^1/2

which I found to be 1.14955 (5dp)
But your answer implies that $x > \sqrt{2}$
$1.14955 < \sqrt{2} \approx 1.414$


$\sin{x}+\cos{x} = \sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right) \implies 0 < \sqrt{\sin{x}+\cos{x}} = \sqrt{\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)} < \sqrt[4]{2} \approx 1.1892...$
 
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