Help with Taylor Series Project

[chongkkmy]

Hi,

My lecture had gave a project about analyzing and discussion about - Taylor Series.

I had done some research and tried understand and solve the question, but I'm in trouble now.
I could only complete No.1 and No.2 (don't know whether is correct or not), I stuck at No.3 I have no idea how to continue, No.4 too. could someone please help me?

 

[HallsofIvy]

You are evaluating the function in (3) incorrectly. You have
f(x)= \sqrt{x} f(4)= \sqrt{x} 4= 4\sqrt{x}
Are you multiplying by 4? f(4)= \sqrt{4}= 2.

Similarly, since
f'(x)= \frac{1}{2\sqrt{x}}, f'(4)= \frac{1}{2\sqrt{4}}= \frac{1}{4}

In my opinion, for (2) you would be better off writing the exact value \sqrt{2}/2 than the approximate "0.70711".

 

[chongkkmy]

Thanks for your reply. I had changed number back to fraction. The answer is more accurate. I had done number 4 as well. I have no idea with number 3, I don't know what to do with the square root. Besides, i would like to ask when differentiate cos X = -sin X thn how a bout -sin X? isn't become -cos X? in differentiation table does not have "-sin X"

 

[Mark44]

Thanks for your reply. I had changed number back to fraction. The answer is more accurate. I had done number 4 as well. I have no idea with number 3, I don't know what to do with the square root.

Can you be clearer on what you're asking? Your function is f(x) = x^1/2. f'(x) = (1/2)x^-1/2, and so on. It's probably easier to calculate your derivatives using exponent notation rather than using radicals. For this problem you need to estimate f(3.8) = f(4 + (-.2)), using a formula similar to what you show in #2.

Besides, i would like to ask when differentiate cos X = -sin X thn how a bout -sin X? isn't become -cos X? in differentiation table does not have "-sin X"

All you need to know for this problem are three rules:

1. d/dx(sin x) = cos x
2. d/dx(cos x) = -sin x
3. d/dx(k f(x)) = k d/dx(f(x))

The third rule can be used when k = -1.

 

[chongkkmy]

Thanks. now I just want to ask about Number 3. I don't know how to start with the "square root".

 

[Mark44]

The first part of my reply was about #3. What part of it don't you understand? \sqrt{x} = x^{1/2}

 

[DMOC]

Edit: nevermind, my comment mirrored mark44, I didn't realize this at first but after a few readings I did.