Homework Statement

Hi there,
I have just started taylor series for my course.. most seems arlgiht so far, however when it comes to validating a given series( tayor or maclaruin), I have an idea on how to find out the x value.. but I don't know what I am doing wrong.

Take for example: The following function (1+x)^-1 . Expand the maclaurin series of it and give values of x for which the above series is valid:

Homework Equations

F(x) = f(0) + f'(0)x /1! + f'' (0)x^2/ 2! + f'''(0)x^3 /3! + f^n (0)x^n/n!

The Attempt at a Solution

Here's my working:

Let f(x) = (1+x)^-1 so f(0) = (1+0)^-1 = 1
f'(x) = -1. (1+x)^-2 f' (0) = -1 . (1)^-2

f''(x) = 2.1 (1+x)^-3 = > f''(0) = 2.1 (1)^-3

f'''(x) = -3.2.1. (1+x)^-4 => f'''(0) = -3.2.1 (1)^-4

Now initially i was puzzled on how to predict (next term or the general pattern) but I think next term will be f'4 = 4.3.2.1 (1x)^-5 so f'4 (0) = 4.3.2.1 (1)^-5

Now I know test ratio can be done to find the radius of convergence i.e for a given function say anx^n ratio would be => (an+1)x^(n+1) / anx^n....

I tried doing the same for the above series.. My general formula is : f(n) : (-1)^n . n! (1+x)^ -(n+1) .... by manipulation i got : 1.(n+1) x [ 1^-n-2] / 1^-n-1]

= > 1. (n+1) but since n => infinity , this function diverges ?
Any constructive input will be appreciated. Thanks!

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Hint: think of geometric series.

Hint: think of geometric series.

I know geometric series goes like : 1+ q+ q^2 +q^3 ... and is in the form of a/1-r but how do I approach this problem ? I am confused .. can't I use ratio method to get value/s of x which converge.

Ok.. so the general form of my series is :

$F^{n}$ = $$(-1)^n (n!) (1+x)^{-(n+1)}$$

So we can ignore $$(-1)^n$$ as n -> infinity.

I am left with : $$(n!) (1+x)^{-(n+1)}$$

$F^{n+1}$ / $F^{n}$= $$(n+1!) (1+x)^{-(n+2)} / n! (1+x)^{-(n+1)}$$

I end up with the expression 1/ (1+x)....

EDIT: I don't mean to spam or brea any forum rules, I can find series and predict nth term.. it's only the convergence bit confusing me :(

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Ok.. so the general form of my series is :

$F^{n}$ = $$(-1)^n (n!) (1+x)^{-(n+1)}$$

How can you express $\frac{1}{1+x}$ as a geometric series? If what you've given above is what you think the nth term of the series is, then it's incorrect. It looks more like $f^{(n)}(x)$, but you can avoid calculating this by using my hint. If you're having trouble, look up geometric series in Wikipedia.

For what values of x does this series converge? I don't think you need to use the ratio test--facts about geometric series should be sufficient.

Well... a /1-r = 1/ x+1 ... I am totally lost...
hm..

I am totally lost...
hm..
If my hint isn't helping you, maybe you should go back to your original approach. Afterward I can show you the faster method I was hinting at if you're interested.

What do you get for $\displaystyle \frac{f^{(n)}(0)}{n!}$?

Coming back to my original approach.. Is the general form of the series wrong ? hm.. I am determined to finish ( and understand it) this problem , tonight.
Thanks for your patience with me.

EDIT: Oh i think i see where i may have gone wrong.
Previous

$F^{n}$ = $$(-1)^n (n!) (1+x)^{-(n+1)}$$

Shouldn't the expression $$(1+x)^{-(n+1)}$$ actually be $$(1+x)^{-n-1}$$ ?
The form which you have posted looks like a general form of maclaurin or taylor series hm...

EDIT: Actually i will give a little read(google) over the 'geometric approach' hm..

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Coming back to my original approach.. Is the general form of the series wrong ?

I think what you had for $f^{(n)}(x)$ was right, but then you tried to do the ratio test using it. The ratio test examines $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n} \right|$ for a power series $\sum_{n=0}^\infty a_n x^n$. So, now that you have $f^{(n)}(x)$, compute $\frac{f^{(n)}(0)}{n!}$, which will give you the coefficient $a_n$.

EDIT: Oh i think i see where i may have gone wrong.
Previous

$F^{n}$ = $$(-1)^n (n!) (1+x)^{-(n+1)}$$

Shouldn't the expression $$(1+x)^{-(n+1)}$$ actually be $$(1+x)^{-n-1}$$ ?
These are the same thing, so both are correct.

The form which you have posted looks like a general form of maclaurin or taylor series hm...

EDIT: Actually i will give a little read(google) over the 'geometric approach' hm..

You can try looking at http://en.wikipedia.org/wiki/Geometric_series" [Broken] link for geometric series. The idea is to find a way to write $\frac{1}{1-x}$ as a geometric series, and then use this to do the same for $\frac{1}{1+x}$.

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Did you get the answer? If you did and still want to see the shortcut using geometric series, I'll post it.

Did you get the answer? If you did and still want to see the shortcut using geometric series, I'll post it.

Sorry other assignments have kept me busy.. i will type down my approach in a bit.If everything fails you can post geometric method.Thanks for your concern =]

Can you show me how to get the co-efficient $a_{n}$ ? (as above).. I am confused.. do I plug in n = 1 ?

Oh do I basically find out my power series and then find out it's general form ? the maclaurin series in this case is : 4

$$1-x+x^2 -x^3 +x^4 -x^5$$

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Can you show me how to get the co-efficient $a_{n}$ ? (as above).. I am confused.. do I plug in n = 1 ?
It's just as you had it above, $a_n = \frac{f^{(n)}(0)}{n!}$.

Oh do I basically find out my power series and then find out it's general form ? the maclaurin series in this case is : 4

$$1-x+x^2 -x^3 +x^4 -x^5$$

That looks like a good start, but you need an infinite series. What is the general pattern? You said before that $f^{(n)}(x) = (-1)^n (n!) (1+x)^{-(n+1)}$. So what is $f^{(n)}(0)$?

It's just as you had it above, $a_n = \frac{f^{(n)}(0)}{n!}$.

That looks like a good start, but you need an infinite series. What is the general pattern? You said before that $f^{(n)}(x) = (-1)^n (n!) (1+x)^{-(n+1)}$. So what is $f^{(n)}(0)$?

I believe I have finally got the solution.. it took me less than a minute to figure it out... the thing which I was over looking at was the fact that I wasn't trying to find out radius of the derivates rather than the series itself. ( As you have pointed out as well).
My series is : $$1-x+x^2 -x^3 +x^4$$
In the general form : $$(-1)^n * x^n$$
This is An.
So the ratio which's in the form of : $a_{n+1}/a{n}$ as n -> infinity becomes:
$$((-1)^n+1) *(x^n+1)/-1^n * x^n$$

after I manipulate them I end up with -x only... which's weird since the limit is given.. how am I suppose to plug in n -> infinity , if n's cancel... hmm

HallsofIvy
Homework Helper
In general, the power series for a function will converge "until there is a problem". Here, the function is 1/(1+ x) which "has a problem" at x= -1. So there is no reason for the series not to converge between 0 and -1 but it obviously cannot converge to a value at x= -1.

Or, as others have pointed out, 1/(1+ x)= 1/(1- (-x)) is the sum of a goemetric series with "common ratio", r= -x. For what "r" does a geometric series converge?

I believe I have finally got the solution.. it took me less than a minute to figure it out... the thing which I was over looking at was the fact that I wasn't trying to find out radius of the derivates rather than the series itself. ( As you have pointed out as well).
My series is : $$1-x+x^2 -x^3 +x^4$$
In the general form : $$(-1)^n * x^n$$
This is An.
Looks good. (You can add the LaTeX markup \cdots to make it $1-x+x^2 -x^3 +x^4 - \cdots$.)
So the ratio which's in the form of : $a_{n+1}/a{n}$ as n -> infinity becomes:
$$((-1)^n+1) *(x^n+1)/-1^n * x^n$$

after I manipulate them I end up with -x only... which's weird since the limit is given.. how am I suppose to plug in n -> infinity , if n's cancel... hmm
That's okay. If $\left| \frac{a_{n+1}}{a_n} \right|$ is the same thing for every n (i.e., doesn't depend on n), then taking the limit shouldn't change anything, right? For example, $\lim_{n \to \infty} 2 = 2$.

Read what the ratio test has to say a bit more carefully. First of all, it examines $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$, so don't forget the absolute values. Now, for which values of L does the series converge? For which does it diverge? And for which is the test inconclusive?

In general, the power series for a function will converge "until there is a problem". Here, the function is 1/(1+ x) which "has a problem" at x= -1. So there is no reason for the series not to converge between 0 and -1 but it obviously cannot converge to a value at x= -1.

Or, as others have pointed out, 1/(1+ x)= 1/(1- (-x)) is the sum of a goemetric series with "common ratio", r= -x. For what "r" does a geometric series converge?

Looks good. (You can add the LaTeX markup \cdots to make it $1-x+x^2 -x^3 +x^4 - \cdots$.)

That's okay. If $\left| \frac{a_{n+1}}{a_n} \right|$ is the same thing for every n (i.e., doesn't depend on n), then taking the limit shouldn't change anything, right? For example, $\lim_{n \to \infty} 2 = 2$.

Read what the ratio test has to say a bit more carefully. First of all, it examines $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$, so don't forget the absolute values. Now, for which values of L does the series converge? For which does it diverge? And for which is the test inconclusive?

Would the r value be r>1 ? since if that's plugged into the equation... i get negative number. I still however am confused since typical ratio test problems have some value of x turning out to be either negative or positive value which tells of divergence,convergence of the function.

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Would the x value be <1 ? since if that's plugged into the equation... i get negative number.
How do you get a negative number since you're taking the absolute value?
I still however am confused since typical ratio test problems have some value of x turning out to be either negative or positive value which tells of divergence,convergence of the function.

Like I said, read carefully what the ratio test tells you: http://en.wikipedia.org/wiki/Ratio_test

How do you get a negative number since you're taking the absolute value?

Like I said, read carefully what the ratio test tells you: http://en.wikipedia.org/wiki/Ratio_test

I see. So in my case L = -x, right ?

I see. So in my case L = -x, right ?

Almost, but don't forget to take the absolute value!

Almost, but don't forget to take the absolute value!

Ah... I see. Thank you very much for your help. I really appreciate it.

So I have another expression: ln (x+a) for which I did the above process all over again.. I end up with L = |x/a|, am I right ?

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Ah... I see. Thank you very much for your help. I really appreciate it.
So what did you get for the radius of convergence?

So I have another expression: ln (x+a) for which I did the above process all over again.. I end up with L = |x/a|, am I right ?

I don't understand... Are you trying to find a series for f(x) = ln(x + a)?

So what did you get for the radius of convergence?

I don't understand... Are you trying to find a series for f(x) = ln(x + a)?
Wouldn't that be simply L = |x| ?

f(x) = ln(x + a)
Series and the x value/s for which it's valid.

Wouldn't that be simply L = |x| ?
Right, and now the link from post #18 tells you for what values of L the series converges. Post what you get for the region of convergence and I'll show you the geometric series trick.

f(x) = ln(x + a)
Series and the x value/s for which it's valid.

Once again, you can either do this problem by calculating successive derivatives, or by noting a nice geometric series trick. What is the derivative of f(x) = ln(x+a) and can you see its connection with the previous problem?

Right, and now the link from post #18 tells you for what values of L the series converges. Post what you get for the region of convergence and I'll show you the geometric series trick.

Once again, you can either do this problem by calculating successive derivatives, or by noting a nice geometric series trick. What is the derivative of f(x) = ln(x+a) and can you see its connection with the previous problem?

Hi there, I have just had my test on this series.... for which i have got 9/10 .. not bad.. for the previous question since nth term's cancel out.. and we're left with -x... for a series to be valid it must converge so I take absolute value of -x i.e |x| .. and the condition under which this series converges is when x<1.

EDIT: In case I am wrong.. I will rethink this again.. I have lecture in few minutes.

for a series to be valid it must converge so I take absolute value of -x i.e |x| .. and the condition under which this series converges is when x<1.
...when |x|<1! Also, remember that when L=1 the test is inconclusive, so you still have to check the endpoints, i.e. determine whether or not the series $\sum_{n=0}^\infty (-1)^n x^n$ converges for $x = \pm 1$.

Okay, you've almost finished the problem, so here's the geometric series trick. We have the equality $\frac{1}{1-r} = \sum_{n=0}^\infty r^n$ for |r|<1. You can prove this equality by noting that the series $(1-r)(1 + r + r^2 + \cdots)$ telescopes and all terms except for the 1 cancel out. Plug in r = -x and we have
$$\frac{1}{1+x} = \frac{1}{1 - (-x)} = \sum_{n=0}^\infty (-x)^n = \sum_{n=0}^\infty (-1)^n x^n$$
which is the same answer you got by computing the derivatives.

For the second problem, my hint is to calculate $\frac{d}{dx} \ln(x+a)$ and see if you can use the first problem to get the answer.