.

**ln(1+x) = x + (x/2)(x/2)(x 2)** .........*in range (0.77,1.54)*

*subtract 0.001 if you are within "5" of end of range (i.e. 0.77 to 0.82, or 1.49 to 1.54)*

*get to range using ln(2) = 0.69315 and ln(5)=1.60944*

.

.

**e^x = 0.1 * [(10 + 10x) + (x^2) * (5 + 2x) ]** .....*in range (0.0,0.5)*

*Get to range using e^0.5 = 1.64872 and e=2.71828*

.

.

Error levels in the ranges are ln(1+x) within 0.0010, and e^x within 0.0013.

Using the Curta calculator to calculate the Natural Logarithm and Exponential functions without using tables!!!

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Natural Logarithm

Preliminary requirements: memorize three numbers.

Ln(2) = 0.69315 (remember as 6-9-3-15)

Ln(5) = 1.60944 (remember as 160-944)

Bottom of range = 0.77

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Step 1.

Eliminate tens

For example, if you want to find ln(187.2), you would find ln(100*1.872), which is equal to ln(100) + ln(1.872)

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Step 2.

Move into the range by eliminating twos or fives.

The range is from 0.77 to 2*0.77, or from 0.77 to 1.54.

In this case, we would say that ln(1.872) = ln(2*0.936) = ln(2) + ln(0.936). Note that 0.936 is in the range [0.77,1.54].

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Step 3.

Use formula for part that is in range. Make sure to subtract 1 first!

Note that the Taylor series expansion for the natural log is as follows:

Ln(1+x) = (x)/1 (x^2)/2 + (x^3)/3 (x^4)/4 + (x^5)/5 for -1 < x < 1

Now a way to show successively improving approximations of this series is as follows:

(x)/1 (x^2)/(1*2)

(x)/1 (x^2)/2 + (x^3)/(2*2)

(x)/1 (x^2)/2 + (x^3)/3 (x^4)/(3*2)

(x)/1 (x^2)/2 + (x^3)/3 (x^4)/4 + (x^5)/(4*2)

We will only need to go as far as the second approximation. It can be rewritten as:

x + (x/2)(x/2)(x 2)

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3A: find x.

Since were using 0.936 = 1+x, we have x = -0.064.

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3B: square half of x.

So to start to do this on the Curta, we will manually enter |x|/2 on the SR. In this case, 0.064 / 2 = 0.032. Square this to get 0.001024. Now enter this in the SR and clear the RR.

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3C: multiply by (x 2)

Multiply this by (-0.064 2) on the Curta. Note that since this multiplication and the addition in step 3D are all negative, we can just do them as positives and add the negative sign later.

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3D: add in x.

Then clear the SR and enter x (-0.064 in this case), and add it in once. Again, since all the additions from 3C and 3D are negative, you can just do them as positives and deal with the negative sign later.

Now you have ln(0.936) = -0.066113536. This will be accurate to within + or - 0.002. The true value is ln(0.936) = -0.0661398.

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Step 4: Add in your twos and fives

We have calculated ln(0.936) = -0.06611. Clear the Curta and put this in the SR, and use one negative turn to subtract it.

Now add in ln(2) = 0.69315, memorized above. Now we get ln(0.936)+ln(2)=0.67204, which means that ln(0.936*2) = ln(1.872) = 0.67204.

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Step 5: Add in your tens.

Since were looking for ln(187.2), we need to further add in ln(2) twice and ln(5) twice. Since we already have ln(2) in the SR, two more positive turns, then two positive turns with ln(5)=1.60944 in the SR.

Final result: ln(187.2) on the Curta is 5.23222, + or 0.002. Actual ln(187.2) is 5.23218. Error is 0.00004.

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Final note on error:

Most of the error of + or 0.002 is in the ends of the [0.77,1.54] range. If you subtract 0.001 if you are very close to the ends, in the [0.77,0.82] and [1.49,1.54] ranges, your error will decrease to + or 0.001.

---

---

Recap of steps:

Step 1. Eliminate tens by division

Step 2. Move into the range [0.77,1.54] by eliminating twos or fives by division.

Step 3. Use formula for part that is in range: ln(1+x) ~= x + (x/2)(x/2)(x 2)

. 3A: find x.

. 3B: square half of x.

. 3C: multiply by (x 2).

. 3D: add in x.

Step 4: Add in your twos and fives

Step 5: Add in your tens.

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*******************

Exponential Function (e^x)

Preliminary requirements: memorize one number

e^0.5 = 1.64872 (remember as 16-48-72)

(assuming you have already memorized e=2.7182818.)

---

Step 1: separate out wholes and halves from remaining fraction.

Exp(1.74) = Exp(1.5) * Exp(0.24)

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Step 2: use formula for fraction

Note that the Taylor series expansion for e^x is

e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + (x^4)/4! + (x^5)/5!

A good approximation for e^x (+ or - 0.0013) over the range [0,0.5] is:

e^x ~= 1 + (x)/1 + (x^2)/2 + (x^3)/5.This can be restated as

0.1 * [(10 + 10x) + (x^2) * (5 + 2x) ]

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2A: find square of x

0.24^2 = 0.0576. Now clear the Curta and put this in the SR.

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2B: multiply by (5 + 2x)

Multiply the 0.0576 by 0.24 twice and by 5 once. You should get 0.315648

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2C: add in (10 + 10x)

Clear the SR, and add in 2.4 once and 10 once. You should get 12.715648.

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2D: Divide by 10 (just move the decimal point).

You will now have 1.2715648

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Step 3: Multiply by whole and half powers of e. Exp(1.74) = Exp(1) * Exp(0.5) * Exp(0.24) = 2.71828 * 1.64872 * 1.27156 = 5.69874

Actual value is 5.69734, error is 0.00140

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Recap of steps:

Step 1: separate out wholes and halves from remaining fraction.

Step 2: use formula for fraction: e^x ~= 0.1 * [(10 + 10x) + (x^2) * (5 + 2x)]

. 2A: find square of x

. 2B: multiply by (5 + 2x)

. 2C: add in (10 + 10x)

. 2D: Divide by 10 (just move the decimal point).

Step 3: Multiply by whole and half powers of e.

********************

Algorithms by Steve A, Curta #34811, May 7, 2004

Edit Text of this page (last edited July 18, 2006 by adsl-69-219-41-94.dsl.chcgil.ameritech.net)

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