Thread: Newton-Raphson

This is the entire problem...

Use 2 repetitions of the Newton-Raphson algorithm to find the value of x near zero for which e^x=2cosx. (answer to 3 decimal places)

What I've done so far...

e^x=2cosx
x=ln(2cosx)
f'(x)=-tanx

xo=0

x1=0-(ln(2cos0)/(-tan0))

.....this is where I get stuck b/c my calculator says ERROR since I can't divide by 0. What am I doing wrong???

Originally Posted by Foxtenn5

This is the entire problem...

Use 2 repetitions of the Newton-Raphson algorithm to find the value of x near zero for which e^x=2cosx. (answer to 3 decimal places)

What I've done so far...

e^x=2cosx
x=ln(2cosx)
f'(x)=-tanx

xo=0

x1=0-(ln(2cos0)/(-tan0))

.....this is where I get stuck b/c my calculator says ERROR since I can't divide by 0. What am I doing wrong???

Do as many methods as you need to to get this



We are trying to find the intersection or in other words the zeros of the difference of the two equations?

So we have

We have out equal to zero...therefore our function is

e^x-2cosx
f'(x)=e^x+2sinx

Xo=0 (?)

X1=0-((e^0-2cos0)/(e^0+2sin0))=1

X2=1-((e^1-2cos1)/(e^1+2sin1))=0.627904


I ended up with 0.628, but this doesn't seem to be the exact zero.

Originally Posted by Foxtenn5

e^x-2cosx
f'(x)=e^x+2sinx

Xo=0 (?)

X1=0-((e^0-2cos0)/(e^0+2sin0))=1

X2=1-((e^1-2cos1)/(e^1+2sin1))=0.627904


I ended up with 0.628, but this doesn't seem to be the exact zero.

Yes...this is a good approximation...that is what this is...just an approximation..the actual answer is .539...but to get that you would have to do a few more iterations...this is good enough for what you were asked to do...good job

Originally Posted by Mathstud28

Do as many methods as you need to to get this



We are trying to find the intersection or in other words the zeros of the difference of the two equations?

So we have

We have out equal to zero...therefore our function is

NR gives the itteration:



Then five iterations give:

Code:

>x0=0
            0 
>x1=x0-(exp(x0)-2*cos(x0))/(exp(x0)+2*sin(x0))
            1 
>x1=x1-(exp(x1)-2*cos(x1))/(exp(x1)+2*sin(x1))
     0.627904 
>x1=x1-(exp(x1)-2*cos(x1))/(exp(x1)+2*sin(x1))
     0.544207 
>x1=x1-(exp(x1)-2*cos(x1))/(exp(x1)+2*sin(x1))
     0.539797 
>x1=x1-(exp(x1)-2*cos(x1))/(exp(x1)+2*sin(x1))
     0.539785

So after 2 iterations you have a relativly poor approximation, you realy need more iterations

RonL