Approximating Limits to Avoid Overflow

Approximating Limits to Avoid Overflow

For most 10-digit scientific calculators, the maximum that a number can be before an overflow error occurs is 9.999999999×10^99.

The Maximum Number

What is the maximum number x can be before the calculator overflows?   (For reference, I used a TI-84 Plus to calculate these numbers.)

Function:  Maximum Number

e^(x^2):   15.17427129

x!:   69    (most calculators only allow nonnegative integers for the factorial function)

3^x:   209.5903274

e^x:  230.2585093

2^x:  332.1928095

x^4:  slightly less than 1E25  (1 × 10^25)  (like 9.99999999999E24)

x^3:  2.15443469E33  (2.15443469 × 10^33)

x^2:  slightly less than 1E50  (1 × 10^50)  (like 9.99999999999E49)

x^(3/2):  4.641588833E66  (4.641588833 × 10^66)  

Limits to Infinity

Why is this important?    This could be useful in applications, say approximating values of y(x) when x approaches infinity.

Example 1:

y = ln(x) / x^3

To find the limit x can be, look at the function presented and select the "part" that has the lowest limit.   In this case, that "part" is x^3.  Approach to that limit.

y(1E33):  7.59853081E-98

y(1.5E33):  2.26343032E-98

y(2E33):  9.58480691E-99

y(2.1E33):  8.28498493E-99

It would appear that ln(x)/x^3 approaches 0 as x approaches ∞.

The key is to select x values that will not cause the calculator to provide the overflow error.

Example 2:  

y = e^x/x^2

e^x has a max x of 230.2585093, while x^2 has a max x of 9.99999999999E49.  To prevent an overflow, we must choose x values approaching the lower max, in this case, 230.2585093.

y(229.9):  1.321975623E95

y(230):  1.459738847E95

y(230.1):  1.611859E95

y(230.2):  1.779832347E95

It would appear that e^x/x^2 approaches ∞ as x approaches ∞.

Example 3:

y = (x^3 - 1) / (x! * x^2)

Since the factorial is involved, the highest x can be is 69.

But y(69) overflows in this case!   Then we have to select lower x values until we can get answers.  This can be a trial and error approach.

y(68):  overflow

y(67):  1.83706434E-93

y(66):  1.21246228E-91

y(65):  7.88100352E-90

Looking at the results from y(65), y(66), and y(67), it appears that (x^3 - 1) / (x! * x^2) approaches 0 as x approaches ∞.

Remember that this isn't proof, but we can get an idea how a rational function will behave.  We still need the tools such as the ratio, root, and other tests.

Eddie

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Calculus Revisited #16: Improper Integrals

Welcome to Part 16 of our 21 part Calculus Revisited series. Today we got down, dirty, and improper.

Improper Integrals


(I)

∞
∫ f(x) dx =
a

lim ( ∫(f(x) dx, a, t) as t → ∞

If f(r) is undefined:

(II)

a
∫ f(x) dx =
r

lim ( ∫f(x) dx, t, a) as t → r+ (t approaches r from the right side)

(III)

r
∫ f(x) dx =
a

lim ( ∫(f(x) dx, a, t) as t → r- (t approaches r from the left side)

Problems

1. Calculate:

4
∫ 2/(x - 2) dx
2

Note 2/(x - 2) is defined at x = 2. This is an improper integral.

∫( 2/(x - 2) dx, 2, 4)
= lim ∫( 2/(x - 2) dx, t, 4) as t → 2+
= lim (2 ln(4 - 2) - 2 ln(t - 2)) as t → 2+

Note: ln t has no limit as t → 0

Therefore the integral has no finite answer. Yes, that can happen.

2. Calculate:

3
∫ dx / √(3 -x)
0

Note that 1/√(3 - x) is undefined at x = 3. Another improper integral.

∫( 1 / √(3 - x) dx, 0, 3)
= lim ∫( 1 / √(3 - x) dx, 0, t) as t → 3-
= lim (- 2 * √(3 - t) + 2 * √(3 - 0) ) as t → 3-
= lim ( -2 * √(3 - t) + 2√3) as t → 3-
= 2√3 ≈ 3.46410

3. Calculate:

∞
∫ e^-x/2 dx
0

So:
∫ (e^(-x/2) dx, 0, ∞)
= lim (e^(-x/2) dx, 0, t) as t → ∞
= lim (-1/2* e^(-t/2) + 1/2 * e^0 ) as t → ∞
= 0 + 1/2
= 1/2

Next time we are going to work with sequences.

Until then, have a great day!

Eddie

This blog is property of Edward Shore. © 2012

Calculus Revisited #4: Limits

Welcome to Calculus Revisited! This is blog entry #4 of a 21 blog entry series. Today, we will cover the basics of limits.

Limit: The limit of a function f(x) is a value of which a function approaches as x creeps closer and closer to it's target x=a. This is not the same as f(a).

In notation:

lim f(x) = L
x→a

The limit may or may not exist.

Left Side Limit: The limit of a function f(x) as x approaches a from the left side. Hence, x < a.

In notation:

lim f(x) = L
x→ a-

Right Side Limit: The limit of a function f(x) as x approaches a from the right side. Hence, x > a.

In notation:

lim f(x) = L
x→ a+

If the left side limit is equal to the right side limit, then the general limit exists at L.

Continuous Function: A function f(x) is continuous if a limit exists for each x in the domain (or specified interval), and that limit is the same as the function's value. It has been often said that you can graph continuous functions without lifting a pencil.

Common continuous functions include:
f(x) = p(x) (polynomials a_n * x^n + ... + a_0)
f(x) = e^x
f(x) = ln x (for x>0 only)
f(x) = sin x
f(x) = cos x

Properties of Limits

lim (f(x) + g(x)) = lim f(x) + lim g(x)

lim c *f(x) = c * lim f(x), c is a constant


Common Ways to Attack Limits

1. If f(x) is continuous, then

lim f(x) = L
x →a

for all x.

2. "Calculator Method": plug in various x_i as x approaches closer and closer to x = a (but not at x = a), observe the results and make a educated conclusion.

3. Graph the function. If graphing calculators are allowed, this is the time to use them.

Problems
1. Find

lim x^2 + 1
x→2

Using the "calculator method":

Left Side Limit: (x approaches 2 with x < 2)
f(2 - .01) = 4.9601
f(2 - .001) = 4.99601
f(2 - .0001) = 4.99960001
f(2 - 10^-9) = 4.999999996
The value is getting close to 5.

Right Side Limit: (x approaches 2 with x > 2)
f(2 + .01) = 5.0401
f(2 + .001) = 5.004001
f(2 + .0001) = 5.00040001
f(2 + 10^-9) = 5.000000004
The value is getting close to 5.

We can reasonably conclude that

lim x^2 + 1 = 5 as x → 2

We could have also observed that x^2 + 1 is continuous everywhere and figured the limit out by plugging in 2 for x.

Now let's go to a case where f(x) is not continuous everywhere.

2. Find

lim 1/(x-2)
x→2

f(2) = 1/0. So plugging in x=2 does not work here.

Left Side Limit: (x approaches 2 with x < 2)
f(2 - .001) = -1,000
f(2 - .0001) = -10,000
f(2 - .00001) = -100,000
f(2 - 10^-9) = -1,000,000,000
Note that f(x) is getting to be a very large negative number, towards -∞

Right Side Limit: (x approaches 2 with x > 2)
f(2 + .001) = 1,000
f(2 + .0001) = 10,000
f(2 + .00001) = 100,000
f(2 + 10^-9) = 1,000,000,000
Note that f(x) is getting to be a very large positive number, towards ∞

But -∞ ≠ ∞

3. Find

lim (sin x)/x
x→ 0

Again, f(0) = 0/0, the plugging it won't work.

Left Side Limit:
f(-.001) = .9999998333
f(-.0001) = .9999999983
f(-.00001) = 1 (calculator returns 1)

Right Side Limit:
f(.001) = .9999998333
f(.0001) = .9999999983
f(.00001) = 1 (calculator returns 1)

From the "Calculator Method":

lim (sin x)/x = 1
x → 0

The next time we will working with derivatives. See you next time, Eddie.

This blog is property of Edward Shore. © 2012