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

vladb

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

vladb

Note that your original formula is wrong, because instead of adding up the terms (with derivatives of various order) you multiply them.
The formula posted by Benorin (4th post) is the correct one, which will give 10 - 1/20 - 1/8000.

To Benorin: < deleted :) >

Last edited: May 23, 2006
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