1. Taylor series problem

Considering the functions,

f(x)=ln(2x)

g
(x)=ln(6x)

(i) By writing
ln(2 x) = ln 2 + ln(1 (x/2) ), and using substitution in one of the standard Taylor series, find the Taylor series about 0 for f. Give explicitly all terms up to the term in x^3. Determine a range of validity for thisTaylor series.

(ii) Use your answer to part (i), and the fact that 6
x = 2 (x4), tofind the Taylor series about 4 for g. Give explicitly the same number of terms as in part (i). Determine a range of validity for this Taylor series.
(iii) Check the first four terms in the Taylor series that you found in part (ii) by finding the first, second and third derivatives of g, and using these to find the cubic Taylor polynomial about 4 for g.

Please I'm struggling to start this exercise I have read my books but I'm not getting how the Taylor series works, maybe is because I still have found a reason for what Taylor sees is used and get bored and distracted every time I work with it.

So an easy and straight forward explanation will also be much appreciated.

xxx

2. Re: Taylor series problem

i) compute the Taylor series for $\ln(1-x)$ and then make the substitution $x\rightarrow \dfrac x 2$ and then add $\ln(2)$ as the "0th" term.

This is a very simple Taylor series you should have no trouble computing.

Once you find the terms of the series apply the ratio test to determine the radius of convergence as usual.

ii) as $\ln(6-x)=\ln(2 - (x-4))$ you can use the series you found in (i) for $\ln(2-x)$ but now make the substitution

$x \rightarrow (x-4)$

iii) should be straightforward.

3. Re: Taylor series problem

With regard to question ii. I get the Taylor series ln6-x/6-x^2/72-x^3/648. I'm doing question iii. I have taken the first deriative of the function ln(6-x) and get 1/x-6. The second derivative of ln(6-x)=-1/(6-x)^2. This isn't the same as the series expansion.

4. Re: Taylor series problem

Originally Posted by michele
Considering the functions,

f(x)=ln(2x)

g
(x)=ln(6x)

(i) By writing
ln(2 x) = ln 2 + ln(1 (x/2) ), and using substitution in one of the standard Taylor series, find the Taylor series about 0 for f. Give explicitly all terms up to the term in x^3. Determine a range of validity for thisTaylor series.

(ii) Use your answer to part (i), and the fact that 6
x = 2 (x4), tofind the Taylor series about 4 for g. Give explicitly the same number of terms as in part (i). Determine a range of validity for this Taylor series.
(iii) Check the first four terms in the Taylor series that you found in part (ii) by finding the first, second and third derivatives of g, and using these to find the cubic Taylor polynomial about 4 for g.

Please I'm struggling to start this exercise I have read my books but I'm not getting how the Taylor series works, maybe is because I still have found a reason for what Taylor sees is used and get bored and distracted every time I work with it.

So an easy and straight forward explanation will also be much appreciated.

xxx
For i) the easiest thing to do is to recall that for x < 1, \displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left[ \ln{ \left( 1 - x \right) } \right] = -\frac{1}{1 - x} \end{align*}, so that means if you can get a series for \displaystyle \begin{align*} \frac{1}{1 - x} \end{align*}, integrate and negate it, you will have a series for \displaystyle \begin{align*} \ln{ \left( 1 - x \right) } \end{align*} which has the same radius of convergence.

As another hint, remember that a geometric series \displaystyle \begin{align*} \sum_{n = 0}^{\infty} r^n = \frac{1}{1 - r} \end{align*} as long as \displaystyle \begin{align*} |r| < 1 \end{align*}. That looks a lot like \displaystyle \begin{align*} \frac{1}{1 - x} \end{align*} to me.

As for your comment about not knowing a reason for what a Taylor Series is used for, well, have a think about your calculator. Do you really think the programmer just magically fitted a button to it so that it could immediately calculate the sine, cosine, tangent, exponential, logarithm, square root, etc of whatever number you put into it? Of course not!

All a calculator really can do is add (and by extension, subtract, multiply, divide and exponentiate - which all boil down to adding anyway). So the question is, if you wanted to say evaluate the logarithm of a number, what is the right combination of additions, subtractions, multiplications, divisions and exponentials to be able to do this?

Now notice, a polynomial is exactly that, a combination of additions, subtractions, multiplications, divisions and exponentials. So by finding a polynomial (and when you have an irrational number, you will need an infinite number of terms) to be able to represent a function makes it possible to know what operations need to be done to that number so that you can evaluate its function value.

THIS is what a Taylor Series is used for.

5. Re: Taylor series problem

Sorry Guys I still don't got it.
Have been trying to solve it but I can move forward I feel so dumb, I need the work-trough with some explanation to help me here.

6. Re: Taylor series problem

I basically gave you the answer. Surely you can replace "r" with "x"...

7. Re: Taylor series problem

Originally Posted by michele
Considering the functions,

f(x)=ln(2x)

g
(x)=ln(6x)

(i) By writing
ln(2 x) = ln 2 + ln(1 (x/2) ), and using substitution in one of the standard Taylor series, find the Taylor series about 0 for f. Give explicitly all terms up to the term in x^3. Determine a range of validity for thisTaylor series.

The Taylor series for the function f(x)=ln⁡(1-x) is given by -x-x^2/2-x^3/3. Substituting x→x/2 the Taylor series is -x/2-x^2/8-x^3/24. The Taylor series for the function f(x)=ln⁡〖(2-x)=ln2+〗-x/2-x^2/8-x^3/24 The validity of the series f(x)=ln⁡(1-x) is |x|<1 The validity of f(x)=ln⁡(1-x)=ln2-log⁡(1-1/2 x)is |1/2 x|<1→|x|

(ii) Use your answer to part (i), and the fact that 6
x = 2 (x4), tofind the Taylor series about 4 for g. Give explicitly the same number of terms as in part (i). Determine a range of validity for this Taylor series

The Taylor series for the function f(x)=ln⁡(6-x) can be rewritten f(x)=ln⁡〖(2-(x-4))〗 substituting x→(x-4) into the Taylor series f(x)=ln⁡〖(2-x)=ln2+〗-(x-4)/2-〖(x-4)〗^2/8-((〖x-4)〗^3)/24 The taylor series of f(x)=ln⁡(6-x) will be valid aboutx-4=0→x=4. The range over which it is valid is |x-4|<a
(iii) Check the first four terms in the Taylor series that you found in part (ii) by finding the first, second and third derivatives of g, and using these to find the cubic Taylor polynomial about 4 for g.

d/dx ln⁡(6-x)=1/(x-6) d^2/(dx^2 ) ln⁡(6-x)=-1/(x-6)^2 d^3/(dx^3 ) ln⁡(6-x)=2/(x-6)^3

I don't know if i'm right. If anyone could help me, that would be appreciated.

xxx
Thanks.

8. Re: Taylor series problem

Hi

Part (I) of the question wants you to use a standard Taylor series. I have put an image of one below.

You will need to take the ln(1-x) form.

To see how ln(2−x) = ln 2 + ln(1−(x/2)) you need to use the following property

of logs => ln(a) + ln(b) = ln(ab) [substituting a = 2 and b = 1−(x/2) ]

Combining gives ln 2 + ln(1−(x/2)) = ln(2(1−(x/2)))

Multiply out the bracket to get ln(2−x) as required.

You can now create a Taylor series substituting x/2 in place of x.

So ln(2−x) = ln 2 + ' your Taylor series using x/2 '

If you want further help with this part or some help on the further parts please let me know.

Regards