# Taylor Polynomial question

1. May 23, 2006

### Math_Frank

Hi Guys,

I have an assigment which I would very much appriciate if You would tell if I have done it correct :)

Use the Taylor Polynomial for $$f(x) = \sqrt(x)$$ of degree 2 in x = 100. To the the approximation for the value $$\sqrt(99)$$

First I find the Taylor polynomial of degree 2.

$$T_2(x) = \sqrt(100) + \frac{\frac{1}{(2) \sqrt(100)}}{1!} (x-100) \frac{\frac{1}{(4) (100)^{3/2}}}{2!} (x-100)^2$$

Last edited: May 23, 2006
2. May 23, 2006

### benorin

now plug-in x=99.

3. May 23, 2006

### Math_Frank

Hello Benorin,

I plugin x = 99 and get

$$T_2(99) = \frac{1599999}{160000}$$

Last edited: May 23, 2006
4. May 23, 2006

### benorin

Only plug-in 99 where you see an x, i.e.

$$T_2(99) = \sqrt(100) + \frac{\frac{1}{(2) \sqrt(100)}}{1!} (99-100) -\frac{\frac{1}{(4) (100)^{3/2}}}{2!} (99-100)^2 = 10+50(1)-\frac{1}{8000}(1)^2=60-\frac{1}{8000}=59.999875$$

5. May 23, 2006

### Math_Frank

The result You get there 59.999875 is that the approximation in procent?

/Frank

6. May 23, 2006

59.99.. obviously cannot be correct. It's not anywhere close to root of 99.

From the formula above I see 10 - 1/20 - 1/8000 = 9.949875. Which squared equals about 99.0000125..., what is pretty close.

7. May 23, 2006

### Math_Frank

Hello Again Benorin,

So the conclusion is that $$T_2(99)$$ cannot be used to find the approximation for 99?

If I insert 99 into the polymial I get $$T_2(99) = 9,999996875$$

which squared gives 99,9999

Is that wrong?

/Frank

Last edited: May 23, 2006
8. May 23, 2006

### benorin

My bad, my post should have read as above, and note that the actual value is closer to

$$\sqrt{99}\approx 9.9498743710661995473447982100121$$

Last edited: May 23, 2006
9. May 23, 2006