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I need a method to calculate ln(x) for small x, other than Taylor series method

by hkBattousai
Tags: converge, logarithm, newton, numeric, taylor
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hkBattousai
#1
Jan26-11, 04:43 AM
P: 58
I'm aiming to calculate ln(x) numerically. I'm using the following procedure for this:
1) If x is greater or equal to 1, use Newton's method.
2) If x is smaller between 0 and 1, use Taylor series expansion.

Newton's method works good, but I have problems with Taylor series expansion method.
[itex]\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} x^n = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \quad{\rm for}\quad \left|x\right| \leq 1\quad[/itex]
Sum of this series is [itex]\small -\infty\normalsize[/itex] for [itex]\small x=-1\normalsize[/itex], but when you try to take sum of it using a computer, it doesn't converge because of computing limitations (like using truncated series). I can't either use Newton's method for this interval, because it has a bad performance for small x values (neither it converges for small x).

For x=0.000001;
ln(x) = -13.8155... (actual)
ln(x) = -1.71828... (calculated)
(The summation goes on until absolute value of any element of the series is below a user defined [itex]\epsilon[/itex] value, which is equal to [itex]1.0 \times 10^{-14}[/itex] for this particular example.)

I need a third method to calculate ln(x) for small x argument.
Can you please suggest me one?
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Pagan Harpoon
#2
Jan26-11, 07:40 AM
P: 85
Maybe you could try directly evaluating the integral:

[tex]\int^{x}_{1}\frac{1}{x}dx[/tex]
hotvette
#3
Jan26-11, 02:09 PM
HW Helper
P: 925
Quote Quote by hkBattousai View Post
I'm aiming to calculate ln(x) numerically. I'm using the following procedure for this:
1) If x is greater or equal to 1, use Newton's method.
2) If x is smaller between 0 and 1, use Taylor series expansion.
What's wrong with Newton's method if x < 1? How are you doing it?

hkBattousai
#4
Jan26-11, 02:12 PM
P: 58
I need a method to calculate ln(x) for small x, other than Taylor series method

Quote Quote by hotvette View Post
What's wrong with Newton's method if x < 1? How are you doing it?
// Public method
long double Math::ln(long double num) throw(MathInvalidArgument)
{
	if (num < 0)
	{
		throw(MathInvalidArgument(L"You tried to take ln() of a negative number."));
	}
	else if (num == 0.0)
	{
		long double zero = 0.0;
		return -1.0 / zero;	// return minus infinity
	}
	else if (num <= 1.0)
	{
		return ln_ByTaylorSeriesExpansion(num);
	}
	else
	{
		return ln_ByNewtonsMethod(num);
	}
}

// Private method
long double Math::ln_ByTaylorSeriesExpansion(long double arg)
{
	arg -= 1.0; // We will use series expansion for ln(x+1)
	long double Result = 0.0, NextSum;
	long double power_of_arg_over_i = 1.0;
	for (uint64_t i=1; true; i++)
	{
		power_of_arg_over_i *= arg / static_cast<long double>(i);
		NextSum = -static_cast<long double>(MinusOneToThePowerOf(i)) * power_of_arg_over_i;
		Result += NextSum;
		if (abs(NextSum) < m_PRECISION_EPSILON)
		{
			break;
		}
	}
	return Result;
}

// Private method
long double Math::ln_ByNewtonsMethod(long double arg)
{
	long double x_k = 1.0, x_k1,asd;
	for (uint64_t i=0; true; i++)
	{
		x_k1 = x_k - ((exp(x_k) - arg) / exp(x_k));
		asd = abs(x_k1 - x_k);	// This value does not approach to zero, fluctuating between 10^(-5) and 10^(-6)
		if (abs(x_k1 - x_k) < m_PRECISION_EPSILON) break;	// m_PRECISION_EPSILON is 10^(-14)
		x_k = x_k1;
	}
	return x_k1;
}
LeonhardEuler
#5
Jan26-11, 03:00 PM
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P: 864
power_of_arg_over_i *= arg / static_cast<long double>(i);
Looking over this quickly, it seems like this isn't what you want. It looks like you're going to end up with a factorial on the bottom since the terms look like they would go

[tex]x,\ \frac{x^2}{2},\ \frac{x^3}{2*3},\ \frac{x^4}{2*3*4} ...[/tex]
LeonhardEuler
#6
Jan26-11, 03:05 PM
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P: 864
Actually, I'm very sure that's the mistake you made since you got
For x=0.000001;
ln(x) = -13.8155... (actual)
ln(x) = -1.71828... (calculated)
If you were calculating what I thought you were calculating, it's the Taylor series for [itex]e^{-x}-1[/itex]. So you would get, after shifting the argument:
[tex]e^{-(0.000001-1)} - 1\approx e-1\approx 1.7182818...[/tex]
hkBattousai
#7
Jan26-11, 03:11 PM
P: 58
Quote Quote by LeonhardEuler View Post
Actually, I'm very sure that's the mistake you made since you got


If you were calculating what I thought you were calculating, it's the Taylor series for [itex]e^{x}-1[/itex]. So you would get:
[tex]e^{0.000001} - 1\approx e-1\approx 1.7182818...[/tex]
Yeah, thanks, there was an error in the code... I feel embarrassed...

The denominator term was was also accumulating into multiplication, so it was becoming factorial of the current for loop variable. It was calculating exp() instead of ln(), what an exciting coincidence...
hkBattousai
#8
Jan26-11, 03:13 PM
P: 58
My new code is:

long double Math::ln(long double num) throw(MathInvalidArgument)
{
	if (num < 0)
	{
		throw(MathInvalidArgument(L"You tried to take ln() of a negative number."));
	}
	else if (num == 0.0)
	{
		long double zero = 0.0;
		return -1.0 / zero;	// return minus infinity
	}
	else if (num <= 1.0)
	{
		return ln_ByTaylorSeriesExpansion(num);
	}
	else
	{
		return ln_ByNewtonsMethod(num);
	}
}

long double Math::ln_ByTaylorSeriesExpansion(long double arg)
{
	arg -= 1.0; // We will use series expansion for ln(x+1)
	long double Result = 0.0, NextSum;
	long double power_of_arg = 1.0, power_of_arg_over_i;
	for (uint64_t i=1; true; i++)
	{
		power_of_arg *= arg;
		power_of_arg_over_i = power_of_arg / static_cast<long double>(i);
		NextSum = -static_cast<long double>(MinusOneToThePowerOf(i)) * power_of_arg_over_i;
		Result += NextSum;
		if (abs(NextSum) < m_PRECISION_EPSILON)
		{
			break;
		}
	}
	return Result;
}

long double Math::ln_ByNewtonsMethod(long double arg)
{
	long double x_k = 1.0, x_k1;
	for (uint64_t i=0; true; i++)
	{
		x_k1 = x_k - ((exp(x_k) - arg) / exp(x_k));
		if (abs(x_k1 - x_k) < m_PRECISION_EPSILON) break;
		x_k = x_k1;
	}
	return x_k1;
}
AlephZero
#9
Jan26-11, 03:13 PM
Engineering
Sci Advisor
HW Helper
Thanks
P: 6,931
When 0 < x < 1, you could use
ln(x) = -ln(1/x)
hkBattousai
#10
Jan26-11, 03:15 PM
P: 58
Quote Quote by AlephZero View Post
When 0 < x < 1, you could use
ln(x) = -ln(1/x)
Very good idea!
This way I can always use Newton's method, it is better since it converges super fast.
Thanks a lot!
hkBattousai
#11
Jan26-11, 03:27 PM
P: 58
Quote Quote by AlephZero View Post
When 0 < x < 1, you could use
ln(x) = -ln(1/x)
Hmm, good advice but numerically not possible to implement.

For example, suppose that you want to calculate
ln(0.000001)
You send it into Newton's algorithm in this form
-ln(1000000)
I need to calculate
exp(1000000)
inside the Newton's algorithm, but obviously it is not possible because of the hardware and software limitations.
And I also realized that, with this code, there is a very low limit for the argument of ln(). I need to work around this problem.
LeonhardEuler
#12
Jan26-11, 03:46 PM
PF Gold
LeonhardEuler's Avatar
P: 864
Quote Quote by hkBattousai View Post
Hmm, good advice but numerically not possible to implement.

For example, suppose that you want to calculate
ln(0.000001)
You send it into Newton's algorithm in this form
-ln(1000000)
I need to calculate
exp(1000000)
inside the Newton's algorithm, but obviously it is not possible because of the hardware and software limitations.
And I also realized that, with this code, there is a very low limit for the argument of ln(). I need to work around this problem.
You could work around it using the property that:
[tex]\ln{(a\cdot 10^{n})}=\ln{a}+n\ln{10}[/tex]
hotvette
#13
Jan26-11, 04:40 PM
HW Helper
P: 925
Hmmm, I must be missing something. I had no problem at all finding ln(0.000001) using hkBattousai's algorithm for Newton's method. It reduces to xk+1 = xk - 1 + arg/exp(xk).


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