How do u solve ((t^2)/2)-cost-1 for 0

Question

Trying to find the value for t which that equation will come out to zero

Answer 1

How do u solve ((t^2)/2)-cost-1 for 0

 

Use Newton-Raphson method.

With the NR method we have the iteration formula,

t_(r+1) = t_r – f(t_r) / f’(t_r)

where f(t) = (t^2)/2 – cos(t) -1

and f‘(t) = t + sin(t).

The function, f(t) is symmetric about t = 0 (for small t < π/2). So let us choose small values of t and see how f(t) varies

t |  0  |    1    |     2    |

y | -2 |-1.04 |1.414 |

There is a change of sign in y for values of t between t = 1 and t = 2. Do it again for smaller increments.

t |   1.2   |   1.4   |  1.6   |

y | -0.64 | -0.19 | 0.31 |

And now our change of sign in y is between t = 1.4 and t = 1.6. So choose our starting iterative value for t as t_0 = 1.5.

 

r             t_r                       f(t_r)                   f’(t-r)                   t_(r+1) = t_r – f(t_r) / f‘(t_r)

0            1.5                       0.0543                 2.4975                 1.4783

1            1.4783                 2.544x10^(-4)    2.4740                 1.47817

2            1.47817              5.777x10^(-9)    2.4739                 1.47817

The last two results (1.47817) are identical to the 5th decimal place.

Accepting this as sufficiently accurate, we can state that our

expression, f(t), is solved for zero when t = 1.47817

Answer 2

You could use Newton's iterative Method and calculus.

First, write f(t)=t²/2-cos(t)-1. Note that t must be measured in radians (not degrees).

Now differentiate: df/dt=f'(t)=t+sin(t).

Newton's Method is:

t_n+1=t_n-f(t_n)/f'(t_n) where t_n is a value close to where we would expect to find t, the solution of the equation f(t)=0. Each iteration, t₁, t₂, etc., should bring us closer to the solution for t.

To get the process started we need t₀. To find such a value close to the solution t, we could sketch a graph or we could just substitute an arbitrary value for t₀. We can't start with t=0 because f'(0)=0 and we would have zero as a denominator (division by zero is not permitted). Let's start with t=1 (radian).

We need a calculator to find each iteration.

t₁=t₀-f(t₀)/f'(t₀)=1-f(1)/f'(1)=1-(0.5-cos(1)-1)/(1+sin(1))=1.56493.

t₂=t₁-f(t₁)/f'(t₁)=1.47969, t₃=1.47817, t₄=1.478170266, t₅=1.4781702664, t₆=1.47817026643.

Depending on how much accuracy we need the process is continued, but we'll take t=1.47817 to be a fairly accurate solution. However, we're not finished yet, because t²=(-t)² and cos(t)=cos(-t), which implies that the full solution has to be t=±1.47817, making two solutions for t (what's true for t must also be true for -t).

A graph would show the two solutions approximately at t=-1.5 and t=1.5. We could have used either of these values for t₀ and arrived at the solution for t much quicker. For example, t₀=1.5, t₁=1.47827, t₂=1.47817.