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Find the solution to this equation using newton-raphson method

I have to find an approximate solution for the equation ln(x)=2-x by substituting suitable values of x.

so x=2-ln(x)
x1=2, x2= 1.306 and i end up with 1.557 as my approximate answer. Then it asks me to do the same using the newton raphson method.

From what i understand thats x2=x1-f(x)/f'(x) where f(x)=2-ln(x) and f'(x) = -1/x

so that means i have x1 - 2-ln(x) / -(1/x)

x-(-(2-ln(x))/1/x)) or x+x(2-ln(x))

but when i use 2 as x1 it converges to a completely different number than the first method.

Have I got the second part wrong? Thanks
Original post by Navm1
...


Using f(x)=2−ln⁡x f(x) = 2-\ln x is incorrect.

Good luck. :smile:
Original post by Navm1
I have to find an approximate solution for the equation ln(x)=2-x by substituting suitable values of x.

so x=2-ln(x)
x1=2, x2= 1.306 and i end up with 1.557 as my approximate answer. Then it asks me to do the same using the newton raphson method.

From what i understand thats x2=x1-f(x)/f'(x) where f(x)=2-ln(x) and f'(x) = -1/x

so that means i have x1 - 2-ln(x) / -(1/x)

x-(-(2-ln(x))/1/x)) or x+x(2-ln(x))

but when i use 2 as x1 it converges to a completely different number than the first method.

Have I got the second part wrong? Thanks


Try f(x)=ln⁡x−2+xf(x)=\ln x -2 + x

(You need it to equal zero)
Reply 3
Avatar for Navm1
Navm1
OP
Original post by Mr M
Try f(x)=ln⁡x−2+xf(x)=\ln x -2 + x

(You need it to equal zero)


i slept on it, dreamt about it then woke up and used 2-lnx-x=0 then ended up using

x1-(-(2-lnx-x/(x+1/x)) and got the answer. thanks for that too
Original post by Navm1
I have to find an approximate solution for the equation ln(x)=2-x by substituting suitable values of x.

so x=2-ln(x)
x1=2, x2= 1.306 and i end up with 1.557 as my approximate answer. Then it asks me to do the same using the newton raphson method.

From what i understand thats x2=x1-f(x)/f'(x) where f(x)=2-ln(x) and f'(x) = -1/x

so that means i have x1 - 2-ln(x) / -(1/x)

x-(-(2-ln(x))/1/x)) or x+x(2-ln(x))

but when i use 2 as x1 it converges to a completely different number than the first method.

Have I got the second part wrong? Thanks


What module is this and what exam board?
Reply 5
Avatar for Navm1
Navm1
OP
Original post by Hazard17
What module is this and what exam board?


Access to HE, think its included in the sequences and series module
Original post by Navm1
I have to find an approximate solution for the equation ln(x)=2-x by substituting suitable values of x.

so x=2-ln(x)
x1=2, x2= 1.306 and i end up with 1.557 as my approximate answer. Then it asks me to do the same using the newton raphson method.

You may be interested to know that Wolfram|Alpha does these.
Reply 7
Avatar for Navm1
Navm1
OP
Original post by Smaug123
You may be interested to know that Wolfram|Alpha does these.


thanks bro. i use wolfram and symbolab all the time but didnt think to try it with newton raphson

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