I was just wondering if anyone can help me understand taylor series for sine and cosine. I have no background in calculus but I always found it interesting how the ratios of the arc to the radius was converted to the ratio of sides of a triangle which has straight lines by using Taylor series. Now I totally understand how to calculate Taylor series I just don’t understand why it works. If someone can enlighten me why we keep adding and subtracting radians to different powers divided by factorials why this creates a more and more accurate number. I’m a visual person so a graph or diagram would be helpful. I don’t have any calculus background so going. Through the function in calculus won’t help much. Please also emphasize why we divide by factorials that to me makes no sense to me. Thank you.
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6If you have not background in calculus it is difficult to understand why it works (you need to understand the proof and that requires a knowledge of the basic calculus concepts such as derivatives and limits). – user Jun 29 '18 at 14:50
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We divide by factorials because that cancels out the effect of the power rule when we inevitably iterate through differentiation. The goal is to make every derivative of the Taylor series match up with every derivative of the target function. – Andrew Li Jun 29 '18 at 14:58
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Related (duplicate?): "Deriving the power series for cosine, using basic geometry". In particular, see my answer. (There's one element of Calculus needed, to guarantee that certain lengths are what we say they are.) – Blue Jun 29 '18 at 15:07
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Just a word to the physicist in you: don’t be worried about taking the square of an angle, and higher powers, too, because $1$ radian is a dimensionless quantity. – Lubin Jun 29 '18 at 16:35
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It is not possible to make it much simpler than what Allawonder stated in his answer. I gave a sketch here of the intuitive motivation behind the exponential function, namely trying to find a function $\exp$ that satisfies $\exp' = \exp$ and $\exp(0) = 1$. This is motivated by the differential equation for exponential decay. Similarly we can intuitively find the power series for $\cos$ and $\sin$, each of which is a solution $f$ to $f'' = -f$, which is motivated by the differential equation for harmonic motion. But intuition is just one step. – user21820 Nov 05 '18 at 09:01
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Saying "intuitively it is true" does not make something more likely to be true. The only way to be sure is to have a logically rigorous proof. In this case, intuition guides us to guess the power series via polynomial approximation (as explained in the link), but it takes non-trivial work to prove that the power series actually converges, and much more work to prove that it has the desired properties! You can prove general theorems about power series first, or you can directly prove it for $\exp$ (properties of $\cos,\sin$ follow easily). – user21820 Nov 05 '18 at 09:08
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I am not aware of any simpler direct proof, and in general you have to realize that mathematics is not a shallow field where you can quickly understand any significant result, because our current mathematical knowledge is essentially refined from hundreds of years of mathematics. – user21820 Nov 05 '18 at 09:13
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Anyway, you're welcome to ask anything on my posts in the basic mathematics chat-room. – user21820 Nov 05 '18 at 09:15
3 Answers
To really understand why a particular mathematical method works, you need to seriously dig into the proofs -- and best as ever, try to construct your own proof. In this case, you need to understand a very famous result known as Taylor's theorem. Lagrange called it the backbone of all calculus (not his exact words, of course, but something along that line, if not more exaggerated). In summary, if a function has derivatives up to some order $n+1$ about any point of an open interval, then it can be approximated about a point of that interval (in a well-defined sense) by the $n$th order Taylor polynomial with an error that can be estimated well.
Moreover, if a function has derivatives of all orders at some point of its domain, and if the sequence of its Taylor polynomials about that point converges to the function, then it may be defined by the Taylor series in some subset of its domain about that point. Such functions are known as analytic.
In any case, if what you really want to know (as I perceive) is how the series was constructed in the first place, I don't know if you can get that as that belongs more to mathematical history than mathematics proper, and some mathematicians don't usually show their results constructively, but in a highly polished, magical form. However, you can still understand this more heuristically (I could go into how I think of this, but I would have to talk in the language of calculus), but you need to understand why polynomials are the simplest of functions in the first place, and to do this you can't do without calculus for long. It's not that hard. If you really do want to understand, then start studying immediately. Resources abound online.
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The rock bottom issue is not the Taylor series, but the notion of "angle". Note that elementary Euclidean geometry does not tell you what the "measure" $\alpha$ of an angle is, but it has all sorts of theorems about sines, cosines, etc. of such "angles" in store. Such theorems can be proven via Pythagoras' theorem, or area considerations (like the sine theorem), but they don't use the measure of angles per se.
Contrasting this the function $t\mapsto\sin t$ is working with the actual "angle measure". It is not possible to understand the Taylor series of this function without entering the "transcendental" world.
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Can anyone give me a recommendation for an online class that will teach me more about this. I find it very interesting . I have a decent trig and algebra background but never took calculus – Physicsrocks Jun 29 '18 at 18:24
Hopefully this will help. Leonhard Euler's power series that evaluates the value of e, a number that is used with complex numbers for many different operations, can be dissected into two separate series (look up the Taylor and Maclaurin series) that give you the sine and cosine functions to whatever accuracy you need.
Note:
= means approximately equal
x^4 means multiply x by itself 4 times
X means multiply
5! means 1 X 2 X 3 X 4 X 5 -- pronounced as "5 factorial"
x^0 = 1 -- (any number raised to the power of 0 is 1)
x^1 = x
e^x -- I'm not going to use this notation because it implies that we are raising e to the power of x, which is not how it really works and causes quite a lot of unnecessary confusion. x in this case, represents the radian measure which measures the circumference then divides this into degrees of 360 as necessary. Instead, I will just treat e^x as you would any other function and use functional notation, calling it e(x) as you would with sin(x). This is not the standard, though.
Euler's power series goes like this...
e(x) = 1 + x + x^2/2! + x^3/3! + x^/4! + ...
You can calculate this on and on to greater accuracy using more terms.
This can be seen also as...
e(x) = 1 + x/1 + (x/1)(x/2) + (x/1)(x/2)(x/3) + (x/1)(x/2)(x/3)(x/4) + ...
That is super cool in my opinion. But wait, there's more...
Let us set x to 1 (radian) and what we get, as we add all the terms, numbers closer and closer to e, the number, which is the base of natural logarithms, 2.7182818284590452.
Now let's separate the terms of the function e(x) into two separate, totally amazing series that give us the sine and cosine formulas in a very elegant way. I've tried to figure out a way to get the sine for a long time and even more so, tried to find the angle for the sine length, or what is called the arcsine. That anyone came up with these formulas is almost beyond my comprehension. Zach Star, Mathologer, 3Blue1Brown (on YouTube)and those who contributed to Wikipedia helped me figure this out, along with others, so credit is due to them and you will want to check them out too.
Check out how the sin(x) and cos(x) functions compare to the e(x) function.
sin(x) = x - x^3/3! + x^5/5! - x^7! + x^9/9! - x^11/11! + ...
cos(x) = 1 - x^2/2! + x^4! - x^6/6! + x^8/8! - x^10/10! + ...
Note:
Sine and cosine only work for angles below PI/2 or 90 degrees and above (-)PI/2. Degrees need to be converted to radians (ie. 72 X PI / 180 = the radian measure for 72 degrees).
The circle has 8 symmetries. On the coordinate plain, as the angle exceeds 45 degrees, the sine and cosine reflect one another. In other words, the sine for 18 degrees equals the cosine for 90 - 18 degrees or 72 degrees and vice versa.
** Both functions have the best approximations when you work with angles in the first 45 degrees or under PI/4 radians.
There are a similar functions for the arcsin and arctan functions, also called inverse functions, that give the angle of a given sine from (-)1 to (+)1 or the give tangent (-)1 < x < (+)1.
Note:
In this case, x is not the radian measure but length of the sine or when looking at the tangent, the ratio of the sine to the cosine, y/x when the radius equals 1.
arcsin(x) = x + (1/2)(x^3/3) + (1/2)(3/4)(x^5/5) (1/2)(3/4)(5/6)(x^7/7) + ...
This function can be simplified into a series of coefficients multiplied by powers of x...
arcsin(x) = x^1 + 0.16666x^3 + 0.075x^5 + 0.0446428571x^7 + 0.0303819444x^9 + 0.0223721591x^11 + 0.0173527641x^13 + 0.0139648438x^15 + 0.0115518009x^17 + 0.0097616095x^19 + 0.0083903358x^21 *** which gives us...
arcsin(x) = 0.78539 useful digits when calculating the square root of 1/2. If you add terms till the x to the power of 41 term, you will get eight digit accuracy.
I have not worked with the arctan(x) function yet.
arctan(x) = x - x^3/3 + x^5/5 - x^7/7 + x^9/9 - x^11/11 + ...
