e^c-2c = e-2
what is c?
ummm, 1?
e^1 = e
2*1 = 2
No! Answer is 0.351
How do I get this?
c = 1 is one solution. To find the other, you can proceed as follows.
e^c-2c - e + 2 = 0
put f(c) = e^c-2c - e + 2
Newton's Method (approximate functon by the tangent) gives succesive better and better approximations.
c_{n+1} = c_{n} - f(c_{n})/f'(c_{n})
= c_{n} -
[e^c_{n}-2c_{n} - e + 2]/[e^c_{n} - 2]
Take c_{1} = 0
then:
c_{2}=0.2817
c_{3}=0.3465
c_{4}= 0.351326
c_{5}= 0.351354757134
c_{6}= 0.351354758153
c_{6} is correct to 12 digits.
To solve the equation e^c-2c = e-2, you can use Newton's method to find an approximate solution. Newton's method involves iterating through an equation to find better and better approximations for the solution.
First, rewrite the equation as f(c) = e^c-2c - e + 2 = 0.
Next, take the derivative of f(c) with respect to c, which is f'(c) = e^c - 2.
Now, let c_{n} represent the nth approximation of the solution.
Apply the Newton's method formula: c_{n+1} = c_{n} - f(c_{n})/f'(c_{n}).
Start with an initial approximation of c_{1} = 0.
Plug in the values into the formula to get the next approximation:
c_{2} = c_{1} - [e^c_{1}-2c_{1} - e + 2]/[e^c_{1} - 2]
Repeat this process iteratively until you reach an approximation that is accurate enough for your needs.
Based on the calculations provided, c_{6} is the approximate solution to the equation, which is approximately 0.351354758153.