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November 9

Definitions of logarithmic and trigonometric functions

What's the definition of sine function? If I wanted to know to what evaluates Sin(50) how could I do it? (Same with logarithm function) 190.60.93.218 (talk) 16:34, 9 November 2012 (UTC)[reply]

I can't do any better than refer you to the article sine. Duoduoduo (talk) 17:03, 9 November 2012 (UTC)[reply]

Which yields the infinite series

sin(x)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}x^{2n+1}

. The input must be in radians. --Melab±1 ☎ 03:10, 11 November 2012 (UTC)[reply]

Also logarithm. And if you just want to find the value of say sin (50), you can just use a technical calculator. Make sure you put it in "deg" mode if you mean 50 degrees, or "rad" mode if you mean 50 radians. Also, the technical calculator will have two different logarithm keys: one called "log" for the log to the base 10, and one called "ln" for the natural log. Duoduoduo (talk) 17:15, 9 November 2012 (UTC)[reply]

I read the article, I think it doesn't show it's function definition, It shows how it uses Taylor series to get an approximation of the value, but I don't think that's the definition, just a way-around. So what's the function definition of Sine? (How would you manually evaluate sine?) (Same for logarithm)190.60.93.218 (talk) 17:21, 9 November 2012 (UTC)[reply]

Page 1 and further of "Real And Complex Analysis" by Walter Rudin. You first define

$\exp(z)\equiv \sum _{n=0}^{\infty }{\frac {z^{n}}{n!}}$

The series coverges absolutely for every z and converges uniformly on every bounded subset of the complex plane. This then proves that exp is a continuous function. And because it converges absolutely, you can multiply exp(z1)*exp(z2) by substituting the series in each factor, do a change in variables in one summation to obtain exp(z1+z2).

Then the following theorem follows:

a) for every z, $\exp(z)\neq 0$

b) exp is its own derivative

c) The restriction of exp to the real axis is a monotonically increasing function, exp(x) tends to infinity for x to infinity and to zero for x to minus infinity.

d) There exists a positive number pi such that exp(i pi/2) = i and such that exp(z) = 1 if and only if z/(2 pi i) is an integer.

e) exp is a periodic function with period 2 pi i.

f) The mapping t to exp(i t) maps the real axis to the unit circle.

g) If w is a complex number and $w\neq 0$ , then w = exp(z) for some z.

The proofs are very simply, I can leave that as a exercise to the OP. For real t, one defines cos(t) and sin(t) to be the real and imagnary parts of the function exp(i t). This fixes all the properties of these functions. Count Iblis (talk) 17:48, 9 November 2012 (UTC)[reply]

(ec)The article sine says In a right triangle, sine gives the ratio of the length of the side opposite to an angle to the length of the hypotenuse. That's the definition -- the sine of e.g. 50° is found by forming a triangle with one 90° angle and one 50° angle. The sine of 50° is the ratio of the length of the side opposite the 50° angle to the length of the hypotenuse (the side opposite the 90° angle). Usually the exact value of the sine is irrational, so you're not going to be able to express it exactly in a finite number of decimal places. But you can get as close as possible by using a Taylor series expansion.

Likewise, the definition of logarithm is given by the article as if $x=b^{y}$ , then y is the logarithm of x to base b, and is written $y=log_{b}(x)$ , so $log_{10}(1000)=3.$ That's the definition: the logarithm is the value of y that solves that equation. To compute it you can use a Taylor expansion to get an arbitrarily close approximation, but again, since it's usually irrational you're never going to be able to express it exactly in a finite number of decimal places. Duoduoduo (talk) 17:58, 9 November 2012 (UTC)[reply]

Here is a possible definition of sine. There is exactly one mapping e from R to C which: 1) is a morphism, i.e., e(x + y) = e(x)e(y); and 2) satisfies e'(0) = i. (In fact, this mapping is e(t) =e^it.) Then sin t is defined as the imaginary part of e(t). The logarithm function can be defined as the inverse of the exponential function exp(t) = e^t from R to (0,+∞). The function exp can be defined like e above, except that its derivative at 0 is to be 1 instead of i.

To evaluate sin 50 (in radians), you might use the Taylor series above. To evaluate log 50, you might find an approximate value for the integral of 1/t from 1 to 50, perhaps by using Simpson's formula.

If you meant sine of 50 degrees, you might let x = 50 degrees, and note that sin 3x = 1/2. Since sin 3x can be expressed easily in terms of sin x via the addition formulas, you can write down a third-degree equation one of whose solutions is sin x. Then use Newton's method, for example, to find the appropriate root. 96.46.202.24 (talk) 10:55, 11 November 2012 (UTC)[reply]

Log(50) = Log(2) + 2 Log(5)

For for $-1<x\leq 1$ , you can use the series expansion:

\log(1+x)=\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {x^{n}}{n}}

This converges very slowly for x = 1, so it can't be used directly. However, for large x, we can write Log(1+x) = Log(x) + Log(1+1/x), which yields:

\log(1+x)=\log(x)+\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {1}{nx^{n}}}

For x = 2 this is still not an improvement, but for x = 5 you obtain a reasonably fast converging series in terms of Log(4)= 2 Log(2). To improve things more, we put $1+{\frac {1}{x}}={\frac {1+y}{1-y}}$ . Then $y={\frac {1}{2x+1}}$ and we have

\log(1+x)=\log(x)+\log(1+y)-\log(1-y)=\log(x)+2\sum _{n=0}^{\infty }{\frac {y^{2n+1}}{2n+1}}

So, we have:

\log(1+x)=\log(x)+2\sum _{n=0}^{\infty }{\frac {1}{(2n+1)(2x+1)^{2n+1}}}

This converges fast for large x, even for x = 1 the convergence is quite fast. The logarithms of the integers can be computed using only fast converging series by exploiting relations such as 3^4 = 81 = 1 + 80. Putting x = 80 in the above expression gives

4\log(3)=4\log(2)+\log(5)+2\sum _{n=0}^{\infty }{\frac {1}{(2n+1)(161)^{2n+1}}}

The relation 7^4 = 1 + 2400 gives:

4\log(7)=5\log(2)+\log(3)+2\log(5)+2\sum _{n=0}^{\infty }{\frac {1}{(2n+1)(4801)^{2n+1}}}

With a few such relations you can solve for the logarithms of the prime numbers. Count Iblis (talk) 18:08, 11 November 2012 (UTC)[reply]

$\log(50)=\sum _{n=0}^{\infty }{\frac {1}{2n+1}}\left(350\left({\frac {1}{99}}\right)^{2n+1}-44\left({\frac {1}{161}}\right)^{2n+1}+156\left({\frac {1}{251}}\right)^{2n+1}+136\left({\frac {1}{4801}}\right)^{2n+1}\right)$

Count Iblis (talk) 22:53, 11 November 2012 (UTC)[reply]

Hello! I would define logarithm as

\ln(x):=\int _{1}^{x}{\frac {1}{t}}dt

.--AnalysisAlgebra (talk) 05:31, 13 November 2012 (UTC)[reply]