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I'm trying to solve the next equation

$$4.68-4.50\cos\alpha-1.23\alpha=0$$

But when using a calculator it gives $\alpha$ as $0.226$ but it s supposed to be $4.483$.

As far as I know it has no closed form solution so its necessary a numeric method, so which is the right number, aside isn't a way to at least manually approximate the solution?

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    $\begingroup$ There is indeed no closed form, but have you tried graphing the equation in DESMOS? $\endgroup$ – imranfat Aug 22 '19 at 1:52
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    $\begingroup$ This is a little baffling. The possible zero(es) near $\alpha = 0$ is/are in the interval $[0.2768, 0.2769]$, which is not close to your $0.226$ and the straightforward zero at $4.5166\dots$ is not close to your $4.483$. Are you sure these numbers and the coefficients in your equation are copied correctly? $\endgroup$ – Eric Towers Aug 22 '19 at 1:58
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    $\begingroup$ I imagine the error between $4.483$ and the true zero might be a result of the intended approximation method, whatever that is - it's close enough to feel that way to me in any event. Like imagine if the exact answer involved an infinite summation, and the exercise intended you to approximate this by taking some finite number of terms in said sum. This is purely a guess though. $\endgroup$ – Eevee Trainer Aug 22 '19 at 2:00
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    $\begingroup$ @EricTowers A closer look shows there is no zero in the interval $[0.2768, 0.2769]$: there is a local minimum at $a = \arcsin(4.50/1.23) \approx 0.2768566278$, where the function value is approximately $0.0108291488$. $\endgroup$ – Robert Israel Aug 22 '19 at 2:01
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    $\begingroup$ @RobertIsrael : Agreed. But the discrepancy between the reported location and the observed location leads me to wonder if there is a transcription error. (I could certainly imagine a calculator's rootfinder getting stuck in the local minimum, but then how does it miss that minimum by so much?) $\endgroup$ – Eric Towers Aug 22 '19 at 2:02
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I don't know how the calculator got its value, $0.226$, or where the value $4.483$ comes from. Plotting $$ f(\alpha) = 4.68 - 4.50 \cos \alpha - 1.23 \alpha $$ gives

f over [-6,10]

We expect this graph to be a cosine with midline given by $4.68 - 1.23 \alpha$, so as soon as a local minimum is above zero, we need not proceed further in that direction (to the left), and as soon as a local maximum is below zero, we need not proceed further in that direction (to the right). The graph above indicates those features, so all possible zeroes are in the interval shown, $\alpha \in [-6,10]$.

Zooming in on the potential zero(es) near $\alpha = 0$, we see that there is a local minimum above the $\alpha$ axis, so there is/are no actual zero(es) there. f over [0,1], with vertical range [0,0.02]

(This plot is an example of why it is useful to include "round" numbers on axes when your interval of interest is very close to that round number. If we only zoom in to show the minimum, we have to actually look at the tick labels to determine whether there is a zero. For instance, whether the above plot shows a zero is immediately clear. Whether the following one does takes a little more inspection.

f on [0.276, 0.278] with vertical range 0.0108305 +/- 0.0000015

)

Going back to the overview plot, there should be a zero of $f$ in the interval $[4,5]$. So let's see if we can find it. (Spoiler: It's at $\alpha = 4.516\,602\,526\,340\,883\,204\,3\dots$.)

Let's separate the polynomial (actually, linear in this problem) and non-polynomial parts and plot them separately. We want $$ 4.68 - 1.23 \alpha = 4.50 \cos \alpha $$ Since $-1 \leq \cos \alpha \leq 1$, the right-hand side is in the interval $[-4.50, 4.50]$. The left-and side is a line and we can solve the the interval of $\alpha$s where the height of the line is in $[-4.50,4.50]$. \begin{align*} -4.50 &\leq 4.68 - 1.23 \alpha \\ -9.18 &\leq -1.23 \alpha \\ 7.463\,414\,634\,146\,341\,463\,4\dots =\frac{9.18}{1.23} &\geq \alpha \\ \end{align*} and \begin{align*} 4.68 - 1.23 \alpha &\leq 4.50 \\ -1.23 \alpha &\leq -0.18 \\ \alpha &\geq \frac{0.18}{1.23} = 0.146\,341\,463\,414\,634\,146\,34\dots \end{align*} (Notice that this possible interval for a zero of $f$ is smaller than the one we got above from looking at the graph.) Plotting the two sides over this interval, we have

Line and cosine graphed together

We know from the above that there is/are no zero(es) of $f$ (intersections of the line from the left-hand side and the cosine from the right-hand side of the above) near $\alpha = 0$, but there is clearly one near $\alpha = 4.5$.

Binary Search

So we can proceed by binary search. We know that the $f$ is positive on one side of the zero and negative on the other, so we can cut the interval containing the zero in half for each evaluation of $f$. You seem to be using thousandths as precision, so we need only evaluate about ten times. We indicate the calculation of $f$ at the midpoint of the current interval containing the zero, then write down the new (smaller) interval which we may infer contains the zero. \begin{align*} f(4) &= 2.7013\dots \\ f(5) &= -2.7464\dots & (4,5) \\ f(4.5) &= 0.09358\dots & (4.5,5) \\ f(4.75) &= -1.3317\dots & (4.5, 4.75) \\ f(4.6) &= -0.4733\dots & (4.5, 4.6) \\ f(4.55) &= -0.1889\dots & (4.5, 4.55) \\ f(4.52) &= -0.01918\dots & (4.5, 4.52) \\ f(4.51) &= 0.03724\dots & (4.51, 4.52) \\ f(4.515) &= 0.009043\dots & (4.515, 4.52) \\ f(4.517) &= -0.002243\dots & (4.515, 4.517) \\ f(4.516) &= 0.003400\dots & (4.516, 4.517) \end{align*} The decimal expansion of every point in that last interval begins "$4.516$", agreeing to three decimal places, so the zero is about $\alpha = 4.516$.

(Note that all we really care about is the sign of these values of $f$. If we know the sign, we know which endpoint to replace with the trial $\alpha$ in that step, so we do not have to evaluate these very carefully (especially in the first few steps). Also, we didn't use the exact midpoints. If we had, the interval containing the zero would be shorter than the interval we found for the same number of evaluations of $f$. It would also need a lot of calculator button pushing since the sequence of midpoints is $4.5$, $4.75$, $4.625$, $4.5625$, $4.53125$, $4.515625$, $4.5234375$, $4.51953125$, $4.517578125$, $4.5166015625$, $4.51708984375$, and $4.516845703125$ to get both endpoints to agree to $3$ decimals.)

Taylor Series

There are other methods we could try. We could replace the cosine with leading segments of its Taylor expansion. Then we are seeking roots of the resulting polynomial. This works if we can center the series close to the root. The Taylor series of cosine centered at $0$ is $$ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots + \frac{x^{2k}}{(2k)!} + \cdots \text{,} $$ but we want to find a zero near $4.5$, so we should center close to $4.5$. Picking standard angles that are around $4.5 \leq 1.5 \cdot \pi$, $\frac{4}{3}\pi = 4.188\dots$ and $\frac{3}{2}\pi = 4.712\dots$, so $\frac{3}{2}\pi$ is closer to $4.5$. The Taylor series of cosine centered at $\frac{3}{2} \pi$ is $$ \cos x = \left( x - \frac{3}{2}\pi \right) - \frac{1}{3!}\left( x - \frac{3}{2}\pi \right)^3 + \frac{1}{5!}\left( x - \frac{3}{2}\pi \right)^5 + \cdots + \frac{1}{(2k+1)!}\left( x - \frac{3}{2}\pi \right)^{2k+1} + \cdots \text{.} $$ We want to use low degrees since polynomials of high degree can be hard to factor. Let's go up to degree $5$ (by $2$s since this series for cosine only has odd degree terms). We list the degree of the initial part of the cosine series we are keeping in that row, the resulting approximation for $f$, and the real roots of that polynomial. \begin{align*} 1& :& -5.73 \alpha + 25.885\dots &= 0 :& \{& 4.517\dots \} \\ 3& :& 0.75 \alpha ^3-10.602\dots \alpha ^2+44.234\dots \alpha -52.598\dots &= 0 :& \{& 2.051\dots, 4.516\dots, 7.569\dots \} \\ 5& :& -0.0375 \alpha ^5+0.883\dots \alpha ^4-7.577\dots \alpha ^3+28.639\dots \alpha ^2-48.227\dots \alpha +34.545 &= 0 :& \{& 4.516\dots\} \end{align*} We get our zero to three decimals almost immediately.

Newton's method

The last method I'll show is Newton's Method. We have prior information that $f$ has a zero near $\alpha = 4$, so we use a linear approximation to $f$ at that point, find that approximation's $\alpha$-intercept, and report that as an improved location of the zero. We need to know that $$ \alpha - \frac{f(\alpha)}{f'(\alpha)} = \frac{150 \alpha \sin \alpha + 150 \cos \alpha - 156}{150 \sin \alpha - 41} \text{.} $$ \begin{align*} 4& :& 4&.5827\dots \\ 4&.5827 :& 4&.5168\dots \\ 4&.5168 :& 4&.5166\dots \\ 4&.5166 :& 4&.5166\dots \end{align*} Since we have found an approximate fixed point of the iteration for Newton's method, and we have graphical evidence that there is a simple zero of $f$ near this last $\alpha$, we have found that $\alpha = 4.516$ is an approximate zero of $f$.

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  • $\begingroup$ Whoah!!!, And I thinked this has nothing left to do about! $\endgroup$ – riccs_0x Aug 23 '19 at 17:25
  • $\begingroup$ And the binary search its new for me! $\endgroup$ – riccs_0x Aug 23 '19 at 18:50
  • $\begingroup$ I think this answer must be locked/protected its highly informative $\endgroup$ – riccs_0x Aug 23 '19 at 19:03
  • $\begingroup$ Binary search in the numerical context is commonly known as bisection method. Conversely, sometimes the binary search in ordered lists is also called bisection. $\endgroup$ – Lutz Lehmann Aug 24 '19 at 8:32
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If you know an approximate of the solution, you can use the fixed-point method to find a less approximate (!) solution. Define $$\alpha_{n+1}=\sqrt[k+1]{{4.68-4.50\cos \alpha_{n}\over 1.23}\cdot \alpha_n^{k}}$$and find the limit. I tried $k=3$ and obtained $\approx 4.5166$. All you have to do is that using a scientific calculator press an approximate of solution e.g. $4$ and then press $Ans$ button. Then input the expression $$\sqrt[k+1]{{4.68-4.50\cos Ans\over 1.23}\cdot Ans^{k}}$$for some $k\in \Bbb R$ and press $"="$ repetitively to obtain a rather precise solution.

P.S.

Such equations are generally impossible to be solved by hand.

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