I can't understand this in partial fractions

[dilan]

Well it say the number = y

and |y|<1

then the product will be less than the number right?

 

[dilan]

I mean when you multiply it by itself.

 

[arildno]

Right!
So when you multiply itself with itself N times, where N is some big number, then $x^{N}$ will be very close to zero,right?

 

[dilan]

Right right it will be very close to zero. I mean it will go on like 0.0000000000001323 like that right?

 

[arildno]

You've got it.

So, if we have $S_{N}=\frac{1-x^{N}}{1-x}$, |x|<1 and N is really big, what will $S_{N}$ be approximately equal to?

 

[dilan]

Will Sn approximately be equal to = 1/1-x

Am I right?

 

[arildno]

Perfectly! (that is 1/(1-x), remember parentheses..)

Thus, it gives perfect meaning to say that as N goes to infinity, Sn converges to a number S=1/(1-x), or that the INFINITE series S is a meaningful concept. Agreed?

 

[dilan]

uhh arildno does converge means like bringing it to one place like? Sorry to ask this because I am from a non-english country?

Ya I agree that now the S has a meaningful value.

 

[dilan]

Oh please continue please? I can really understand what you teach me than my school teacher.

 

[arildno]

So, what we have found out, is that the INFINITE series,
$$S=\sum_{n=0}^{\infty}x^{n}$$
is a meaningful concept, as long as |x|<1.
We call 1 here to be the radius of convergence for the infinite series, that is the bound we must put on x, in order for the infinite series to have any meaning (I.e, being some number).
Okay?

 

[dilan]

Ya right. 1 is the bound because if it gose beyond 1 then we won't find a bound because we can get large numbers right?

Anyway this is the first time I used this symbol next to the "Sn ="
Can you teach me about this also?
$$S=\sum_{n=0}^{\infty}x^{n}$$

 

[VietDao29]

Ya right. 1 is the bound because if it gose beyond 1 then we won't find a bound because we can get large numbers right?

Anyway this is the first time I used this symbol next to the "Sn ="
Can you teach me about this also?
$$S=\sum_{n=0}^{\infty}x^{n}$$

$$\sum_{i = m} ^ n$$, this is the summation symbol. It's a capital Sigma.
i represents the index of summation; m is the lower bound of summation, and n is the upper bound of summation.
----------
$$\sum_{i = m} ^ n (a_i)$$
means that all you need is just to sum from a_m to a_n
$$\sum_{i = m} ^ n (a_i) = a_m + a_{m + 1} + a_{m + 2} + ... + a_{n - 1} + a_n$$
----------------
Example:
$$\sum_{k = 1} ^ 3 \left( \frac{k}{k + 1} \right) = \frac{1}{1 + 1} + \frac{2}{2 + 1} + \frac{3}{3 + 1} = \frac{23}{12}$$
----------------
For Convergent series, you may want to have a look here: Convergent series.
If you say some series $$S_N = \sum_{k = 1} ^ N (a_k)$$ converges to some number L, then it means that: $$\lim_{N \rightarrow \infty} S_N = L, \ N \in \mathbb{N}$$
Can you get this? :)

 

[dilan]

Ok I got it now

So, what we have found out, is that the INFINITE series,
$$S=\sum_{n=0}^{\infty}x^{n}$$
is a meaningful concept, as long as |x|<1.
We call 1 here to be the radius of convergence for the infinite series, that is the bound we must put on x, in order for the infinite series to have any meaning (I.e, being some number).
Okay?

Hi VietDao29 thanks for your post about the meaning of the Capital letter of zigma with the other letters. Realy useful. Thanks alot.

Ya now I undertand. Okay we have now got a meaningful concept for Sn. Earlier I had a little problem in undertanding the symbol, but now I get it. Okay now we have a meaningful concept as long as |x|<1

But I don't know anything about this
$$\lim_{N \rightarrow \infty} S_N = L, \ N \in \mathbb{N}$$

Anyway arildno I hope that this will not be needed for now. If yes I would like to know about that also.
Okay upto here now I can get it. What's the next step?

 

[VietDao29]

But I don't know anything about this
$$\lim_{N \rightarrow \infty} S_N = L, \ N \in \mathbb{N}$$

Have you learned limit by the way? Something looks like:
$$\lim_{x \rightarrow 3} x ^ 2 = 9$$ (this is an example of limit of a function)
or:
$$\lim_{n \rightarrow \infty} \frac{1}{n} = 0$$ (an example of limit of a sequence)?

 

[arildno]

Okay, just substitute S for L, and the statement $\lim_{N\to\infty}S_{N}=S$ means that the value of $S_{N}$ approaches S as N trundles off into infinity. (Something you already know).
$N\in\mathbb{N}$ just means that N is a natural number (1, 2, 350000)and so on (not a fraction or decimal number).

 

[dilan]

Oh I see. So this $\lim_{N\to\infty}S_{N}=S$ is something about a limit. Realy great to learn about this. I didn't know about this. Actualy I haven't done this yet. Ok I get it now.

ok arildno I get it upto here without any problem. Now my next step will be?

 

[arildno]

Okay, you said that resolving into partial fractions is no problem, right?

 

[dilan]

Ya no problem at all.

 

[arildno]

All right, then!
Now, consider a fraction of the form $\frac{1}{1+2x}$
If we write this as:
[tex]\frac{1}{1+2x}=\frac{1}{1-(-2x)}[/itex]
then from the above, we can write:
$$\frac{1}{1-(-2x)}=\sum_{n=0}^{\infty}(-2x)^{n}$$
as long as |-2x|<1, that is |x|<1/2 (Remember that a number x and its negative, -x, has the same ABSOLUTE value!)

Agreed?

 

[dilan]

Yep agreed

 

[arildno]

Furthermore, if you are to find the coefficient for the fourth power of x in this series expansion of 1/(1+2x), then that would be found from the term $(-2x)^{4}$, i.e, the coefficient is $(-2)^{4}=16$
Agreed?

 

[dilan]

oh you I see. Yep agreed.

 

[arildno]

Okay, suppose we are to find the coefficient for the fourth power of the series expansion of the fraction $\frac{1}{(1+2x)^{2}}$
As long as |x|<1/2, we may write:
$$\frac{1}{(1+2x)^{2}}=\frac{1}{1+2x}*\frac{1}{1+2x}=(\sum_{n=0}^{\infty}(-2x)^{n})*(\sum_{n=0}^{\infty}(-2x)^{n})$$

Now, as we are only interested in determining the coefficients to the fourth power of x, then we need only care about those products of terms from the first series with terms from the second series that yield something with $x^{4}$

In particular, those products will be:
1. Multiplying the 0th power term in the first series with the 4th power term in the second series
2.Multiplying the 1st power term in the first series with the 3rd power term in the second series
3.Multiplying the 2nd power term in the first series with the 2nd power term in the second series
4.Multiplying the 3rd power term in the first series with the 1st power term in the second series
5.Multiplying the 4th power term in the first series with the 0th power term in the second

That is, we will find the coefficient of the 4th power of x in the series product by calculating:
$$(-2x)^{0}*(-2x)^{4}+(-2x)^{1}*(-2x)^{3}+(-2x)^{2}*(-2x)^{2}+(-2x)^{3}*(-2x)+(-2x)^{4}*(-2x)^{0}=5*16x^{4}=80x^{4}$$

That is, the coefficent is 80.
Agreed?

 

[dilan]

Is this like the pascal trangle. I mean some what related to the binomial theorem?

 

[arildno]

Somewhat similar to that, yes!

But, the main thing is, now you know how to find the power series representation of any fraction of the form 1/(1-ax), and also, how to find the coefficient of a particular power for products like 1/(1-ax)^2
That's basically all you need in order to answer your original question.
Finally, remember that EVERY power series you make must be within its own radius of convergence in order to be meaningful; thus, the interval for which ALL your power series are valid on is that which is less than the LEAST radius of convergence.

 

[arildno]

Hmm..I just reviewed your original question:
It seems to want the general termof the expansion, rather than the specific one.

1.Have you learned about the Cauchy product of series yet?

2. Have you learned about Taylorseries, and how to compute them?

 

[dilan]

you mean like |x|<1 right?

 

[dilan]

I am sort of

Hmm..I just reviewed your original question:
It seems to want the general termof the expansion, rather than the specific one.

1.Have you learned about the Cauchy product of series yet?

2. Have you learned about Taylorseries, and how to compute them?

Sorry arildno I havn't learned any of them. Will it be very difficult for me to learn it?

 

[arildno]

What sort of class are you taking?
I'm suddenly unsure as to what method your teacher had in mind when giving that exercise.

 

[dilan]

I am just in the A/L class. Just finished my Local O/Ls. I just swiched to the english medium now. I don't know, really that teacher is too advance sort of. Even the other students also sort of get fedup of maths like. But I am trying to somehow keep up. He told us that there is a big gap between O/Ls and A/Ls. He told that it is very difficult to cover this gap. He thinks that we have done everything. I don't know whether in the London sylluba it's there. But in our O/L syllubas non of this is there that you thought above except for the equation

Sn = n/2 (a+l)
and
Sn = n/2 (2a + (n-1)d)

Well that's the way. In our country only from arround 20% - #8% like pass Maths.
I am just trying to come into the standard. Well I believe that PF was the best place for me.

 

[arildno]

Seems that you've got a tough time ahead, but the right attitude to face it then!

Now, have you learned that a function f(x) may be written in series form:
[itex]f(x)=f(0)+f'(0)x+f''(0)\frac{x^{2}}{1*2}+f'''(0)\frac{x^{3}}{1*2*3}+++[/tex]
where, say, f''(0) means the 2nd derivative of f, evaluated at x=0?

 

[dilan]

Well

Seems that you've got a tough time ahead, but the right attitude to face it then!

Now, have you learned that a function f(x) may be written in series form:
[itex]f(x)=f(0)+f'(0)x+f''(0)\frac{x^{2}}{1*2}+f'''(0)\frac{x^{3}}{1*2*3}+++[/tex]
where, say, f''(0) means the 2nd derivative of f, evaluated at x=0?

I am sort of like stuck. But I have to face these. TO be honest arildno I don't know that also.
What am I to do? have any website that I can learn all these and come back to you?

 

[arildno]

Well, it isn't as bad as you think, because what I've tried to find out was if there was an ALTERNATIVE method of solving your problem than the one I initially thought of, and that you knew about.
That is; It isn't loads and loads of new stuff you have to learn in order to tackle your original question, from what we've covered in this thread, we may well find a way to answer your question without using the concept of derivatives, Taylor series and so on.

Just give me a bit of time, OK?

 

[dilan]

Ok no problem. Realy appreciate your interest in helping me. Can't believe, but really you are great person.
Thanks alot
Hope I will find a solution.

 

[dilan]

hi arildno,

Uhh were you able to get anything? Well it's ok take your time, I am not in a hurry.