# Homework Help: Help (Taylor Series)

1. Oct 1, 2009

### chongkkmy

Hi,

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?

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2. Oct 1, 2009

### 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".

3. Oct 1, 2009

### 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"

4. Oct 1, 2009

### Staff: Mentor

Can you be clearer on what you're asking? Your function is f(x) = x1/2. 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.
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.

5. Oct 1, 2009

6. Oct 1, 2009

### Staff: Mentor

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

7. Oct 1, 2009

### DMOC

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