Linear Approximation

Computer Science Level 2

Let's say want to computer the value of a function $f(x)$ at $x = 1$ but all we know about this function is this:

$f(x) = f'(x)$

and

$f(0) = 1$

One way to do this to use linear approximation repeatedly with a defined step size (say h). The lesser will be the value of step, the better approximation we can have.

Let's take an example,

Let's use the step size of $0.5$

Using, a step size of $0.5$ or $h = 0.5$ , we can first compute the value at $f(0.5)$ and then at f(1) which we ultimately require.

Generally, linear approximation states:

$f(h) \approx f(0) + h.f'(0)$

$f(2h) \approx f(h) + h.f'(h)$

$f(3h) \approx f(2h) + h.f'(2h)$

and this can go forever.

To approximate the value of f(1) using a step size of $0.5$ that is for $h = 0.5$ , we can do it like this:

First, we compute value of $f$ at $0.5$ like this:

$f(0.5)\approx f(0) + 0.5 \times f'( 0) = 1 + 0.5 \times 1 = 1.5$

Secondly, we can use the value of $f$ at $0.5$ to computer $f(1)$ like this:

$f(1) \approx f(0.5) + 0.5 \times f'(0.5) = 1.5 + 0.5 \times 1.5 = 2.25$

Now, write a program to find the value of $f(1)$ using a step size of $0.05$ or $h = 0.05$ with repeated linear approximations.

The approximation of $f(1)$ using this program is:

2.673 2.663 2.663 2.653

2 solutions

Michael Mendrin
Apr 3, 2014

f(20) is just (1+0.05)^20 = 2.6532977051444213..., no need to write a program. Indeed, as the step size -> 0, this value becomes e, since (1 + 1/n)^n = e for as n -> ∞

Of course, for other functions, it isn't as simple as this.

Tried to write a program in Java......but i am getting a really huge value as result.....what should be the correct programming logic for this?

Vaishnavi Gupta - 7 years, 2 months ago

Lokesh Sharma
Apr 1, 2014

Solution using Python:

def linearApprox(h):
    '''
    h: Step size for repeated linear approximation
        Returns: Approximated value of f(1) 
        '''
    res = 1

    for i in range(int(1/h)): # Runs the loop n number of times such that h*n = 1
        res = res + h*res

    return res

print linearApprox(0.05)

Hahaha, Lokesh, I don't know the solution but the pic is really good... Cheers!!!

A Former Brilliant Member - 7 years, 2 months ago

Yeah, I actually created the question just to show this picture. Cheers!!!

Lokesh Sharma - 7 years, 2 months ago

Linear Approximation

2 solutions

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