Taylor Polynomial for Square Root Function at x = 100

[Math_Frank]

Hi Guys,

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

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

First I find the Taylor polynomial of degree 2.

[tex]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[/tex]

 

[benorin]

now plug-in x=99.

 

[Math_Frank]

now plug-in x=99.

Hello Benorin,

I plugin x = 99 and get

[tex]T_2(99) = \frac{1599999}{160000}[/tex]

 

[benorin]

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

[tex]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[/tex]

 

[Math_Frank]

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

[tex]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[/tex]

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

/Frank

 

[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.

 

[Math_Frank]

Hello Again Benorin,

So the conclusion is that [tex]T_2(99)[/tex] cannot be used to find the approximation for 99?

If I insert 99 into the polymial I get [tex]T_2(99) = 9,999996875[/tex]

which squared gives 99,9999

Is that wrong?

/Frank

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.

 

[benorin]

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

[tex]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+\frac{1}{20}(-1)-\frac{1}{8000}(1)^2=10-\frac{1}{20}-\frac{1}{8000}=9.949875[/tex]

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

[tex]\sqrt{99}\approx 9.9498743710661995473447982100121[/tex]

 

[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 :) >