• Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

how to expand the binomial (x^2 + y^2)^(3/2)

Status
Not open for further replies.

PG1995

Active Member
Hi

Could you please tell me how to expand the following binomial expression?



Thank you.

Regards
PG
 

MrAl

Well-Known Member
Most Helpful Member
Hi,

Do you mean you want:

[LATEX]\sqrt{y^6+3 x^2 y^4+3 x^4 y^2+x^6}[/LATEX]

or

[LATEX](x^2+y^2)\sqrt{x^2+y^2}[/LATEX]

?
 
Last edited:

PG1995

Active Member
Hi,

Do you mean you want:

[LATEX]\sqrt{y^6+3 x^2 y^4+3 x^4 y^2+x^6}[/LATEX]

or

[LATEX](x^2+y^2)\sqrt{x^2+y^2}[/LATEX]

?
Hello MrAl :)

The first one: [LATEX]\sqrt{y^6+3 x^2 y^4+3 x^4 y^2+x^6}[/LATEX]

By the way, do you have any clue that how one can expand the expressions involving fractional exponents such as {x^2+y^2}^(1/2) on the calculators such TI-89?

Thank you.

Best wishes
PG
 

MrAl

Well-Known Member
Most Helpful Member
Hi,

Well, you dont really 'expand' that in the same way you dont expand sqrt(x). Not everything can be reduced or changed. If you can accept complex numbers in your application you can factor the 'inside' like this:

[LATEX] \sqrt{(y-i x)(y+i x)}[/LATEX]

or:

[LATEX]\sqrt{(x-i y)(x+i y)}[/LATEX]

if that does you any good, and then we also have:

[LATEX]\sqrt{a b}=\sqrt{a}\sqrt{b}[/LATEX]

so we can get:

[LATEX]\sqrt{y-ix}\sqrt{y+ix}[/LATEX]

or:

[LATEX]\sqrt{x-iy}\sqrt{x+iy}[/LATEX]

if that helps at all.

Sometimes we also have:

[LATEX]R^2=x^2+y^2[/LATEX]

taking the square root of both sides we get:

[LATEX]R=\pm(x^2+y^2)^\frac{1}{2}[/LATEX]

and taking that up to the third power we get:

[LATEX]R^3=\pm(x^2+y^2)^\frac{3}{2}[/LATEX]

so that whole right side is equal to [LATEX]R^3[/LATEX].
 
Last edited:

PG1995

Active Member
Thanks a lot, MrAl.

By the way, I just figured out how to do it on the calculator. I wanted to do it on the calculator because it was not directly part of my work. Please see the attachment.

Best wishes
PG
 

MrAl

Well-Known Member
Most Helpful Member
Hello,

From your last post i guess i dont know what you are after here. For example, why did you want to take the limit as y goes to zero?
Also, i think the result is plus or minus but im not sure what you want to do with that anyway.

When you say 'expand', what are you looking for?
 

PG1995

Active Member
Hi

I was just curious to see how it is expanded the way (x+y)^2 is expanded into x^+y^2+2xy. Originally, I only wanted to take the limit and I could have done it directly without any expansion! Are you implying that the expansion result produced by the calculator needs plus/minus sign? Or, perhaps you are saying that the limit result should have plus/minus sign. I believe this is the case. Yes, when I take the limit directly it gives absolute value for the result. Thank you.






Regards
PG
 
Last edited:

MrAl

Well-Known Member
Most Helpful Member
Hi,

So your original intent was to take the limit as y goes to zero, ok.

Yes the absolute value sign is needed because when x goes negative the result is still positive, and as we all know when we take a negative number up to an odd power we get a negative number back again. So either we have to state the result as x^3 for positive x and -x^3 for negative x, or as abs(x^3) or abs(x)^3.
 

Ratchit

Well-Known Member
PG1995,

Could you please tell me how to expand the following binomial expression?

http://www.electro-tech-online.com/mathematics-physics/mathematics-physics/a...1&d=1346407856
The expansion of a binomial expression is a straight forward operation. Why are you having trouble with it? Because the exponent is not a positive integer, the expansion will be an infinite series. In the attachment is the first six terms of the infinite series. Included is a comparison of the exact value and the first six terms of the binomial expansion.

Ratch
 
Last edited:

MrAl

Well-Known Member
Most Helpful Member
PG1995,



The expansion of a binomial expression is a straight forward operation. Why are you having trouble with it? Because the exponent is not a positive integer, the expansion will be an infinite series. In the attachment is the first six terms of the infinite series. Included is a comparison of the exact value and the first six terms of the binomial expansion.

Ratch

Hi,


I cant help but wonder what you intended to do with this equation, what you wanted to use it for.

It's also pretty misleading to show solutions that match for a well chosen set of x and y alone. The equation isnt even symmetrical in x and y as the original surely is.

For example, chose x=1 and y=2 and the results vary from the original expression by some 400 percent so they are not even remotely close.
 

Ratchit

Well-Known Member
MrAl,

I cant help but wonder what you intended to do with this equation, what you wanted to use it for.
By equation, I assume you mean the binomial expansion? I don't have any use for it, except to illustrate to PG how to expand a binomial expression.

It's also pretty misleading to show solutions that match for a well chosen set of x and y alone. The equation isnt even symmetrical in x and y as the original surely is.
The full expansion is a infinite series, so the last terms will complete the symmetry.

For example, chose x=1 and y=2 and the results vary from the original expression by some 400 percent so they are not even remotely close.
What you say is true, because the values you chose make the series diverge. Now, if you chose x=2, y=1, the radicand will not change, and the series will converge. So you want to chose x to be as large as possible compared to y to make the series converge faster.

Ratch
 

MrAl

Well-Known Member
Most Helpful Member
Hello,


Expanding the binomial in this way doesnt seem to do any good, i mean accomplish any purpose other than to show how an expansion might be done, which doesnt apply to this thread even though it has interesting side lines :)

With an expansion i dont want to be forced to carefully choose my x and y, there are better expansions that are not so picky. It makes sense that any expansion should definitely include x=1 as it seems very silly to not be able to choose that value or close to that value.

But going to more terms doesnt seem to help either with the problem with the selection of x. I am thinking that there may be a limit that has to be imposed (you might check into this), but it may actually get worse with more terms. This could be due to numerical instability in the normal computer floating point unit which is limited in precision but i havent investigated this.

But in showing the expansion i think you did a good job of illustrating that. What would be interesting would be to see a proof of the binomial expansion with non integer powers. But then again the practicality may be severely limited do to the aforementioned CPU floating point limitations even in the modern computer, and high precision numerical routines will slow things down quite a bit.

I can think of one expansion that works pretty nice for square root. It involves taking the natural log however (but only once no matter how many terms in the expansion), but that can be achieved with another expansion. It's very stable over a very wide range of x too that goes right down to 0^+, the right side of zero.
 
Last edited:

PG1995

Active Member
Thank you, Ratch.

It wasn't simple to me but now I have looked it up carefully. I believe you have used this formula:



Best wishes
PG
 
Last edited:

Ratchit

Well-Known Member
MrAl,

Expanding the binomial in this way doesnt seem to do any good,...
It showed PG how to expand a binomial. Its uses include calculating an approximate value.

i mean accomplish any purpose other than to show how an expansion might be done,
I don't know how to expand a binomial any other way.

which doesnt apply to this thread even though it has interesting side lines :)
It answered PG's question.

With an expansion i dont want to be forced to carefully choose my x and y, there are better expansions that are not so picky.
A binomial expansion is what it is. You have to live with it.

It makes sense that any expansion should definitely include x=1 as it seems very silly to not be able to choose that value or close to that value.
You will get convergence anytime x > y in (x+y)^n .

But going to more terms doesnt seem to help either with the problem with the selection of x.
The selection of x or y deals with convergence, not the number of terms.

I am thinking that there may be a limit that has to be imposed (you might check into this), but it may actually get worse with more terms.
If convergent, a greater number of terms bring the approximation closer to the true value.

This could be due to numerical instability in the normal computer floating point unit which is limited in precision but i havent investigated this.
How you calculate high precision is up to you.

But in showing the expansion i think you did a good job of illustrating that. What would be interesting would be to see a proof of the binomial expansion with non integer powers. But then again the practicality may be severely limited do to the aforementioned CPU floating point limitations even in the modern computer, and high precision numerical routines will slow things down quite a bit.
Thank you. You can Google for the proof. Like I said, how you do high precision is up to you.

I can think of one expansion that works pretty nice for square root. It involves taking the natural log however (but only once no matter how many terms in the expansion), but that can be achieved with another expansion. It's very stable over a very wide range of x too that goes right down to 0^+, the right side of zero.
There are a lot of ways to do the same thing.

Ratch
 

MrAl

Well-Known Member
Most Helpful Member
Thank you, Ratch.

It wasn't simple to me but now I have looked it up carefully. I believe you have used this formula:



Best wishes
PG

[Note: convergence test far below]

Hello again,

You have to be a little careful here how you state the formula. When you use the (n/k) form you'll get negative numbers which strictly speaking dont work inside the factorial. Your first representation is the preferred form as that doent have the same problem.

You also have to be very careful with the domain as i tried to tell Ratch. Certain x and y will give totally inaccurate results being off by 400 percent or more (a result of say 11 when the true result is 2 or something like that), and this problem doesnt seem to get any better with an increased number of terms (and may actually get worse). There are much better expansions that dont have this problem based on other math ideas and im sure they are illustrated on the web.

The reason i was talking about the theory of the expansion was because i would bet there is a limiting value for the domain when we expand like this. It would be nice to know how to calculate the domain since the expansion is so sensitive to the selection of x and y.

Whenever an expansion is done like this (meaning any method not just this one) there is always a domain statement that goes with it. We just dont blindly apply some formula and hope that it works. We've all seen this like in:
y=sqrt(x), where 0 <= x < +inf
as a simple example. Approximations are known to be inaccurate for certain domains so that's why the domain always accompanies the approximation. Without the statement of the domain we dont know if we can apply that particular formula or not.
There are sometimes multiple domain statements too where the limit of the precision is stated with a given domain. For example:
+/- 1 percent: -2 < x < 2
+/- 0.1 percent: -1/3 < x < 1/3
etc., which here would indicate that as x gets larger the accuracy falls but for small x it works pretty well.

ADDED LATER:
Ok i did a quick convergence test on the expansion we had been talking about previously and found that it converges when:

[LATEX] \frac{\mid y \mid}{|x|}<1[/LATEX]

This means the results will start to get inaccurate when the x and y do not meet this criterion. This kind of thing is typical of non integer expansions but i did do this a little fast so check it over when you feel like it.
Note this is when in the above expansion formula a=x^2 and the x is really y^2 so we do (x^2+y^2) in that order.
 
Last edited:

PG1995

Active Member
Thank you, Ratch, MrAl.

Ratchit said:
So you want to chose x to be as large as possible compared to y to make the series converge faster.
The way I have written the series x=a and y=x. Yes, "a" should be as large as possible compared to x to make the series converge faster. For instance, in the highlighted row, there is a difference of 10,000 between the exact value and approximated value.





@MrAl: I just saw your post. Thank you. Once I have read it carefully I will let you know if I have any queries. Thanks.

Regards
PG
 
Last edited:

MrAl

Well-Known Member
Most Helpful Member
Thank you, Ratch, MrAl.



The way I have written the series x=a and y=x. Yes, "a" should be as large as possible compared to x to make the series converge faster. For instance, in the highlighted row, there is a difference of 10,000 between the exact value and approximated value.





@MrAl: I just saw your post. Thank you. Once I have read it carefully I will let you know if I have any queries. Thanks.

Regards
PG

Hi PG,

Ok sure. So you did a different expansion this time? It's good to see some decent range of results as in your table. In this case convergence will not be defined by
abs(y)/abs(x)<1

as it was before, but will be now:
abs(x)/abs(a)<1

Note the difference.

Wait a minute...
Are you saying you did (a+x)^5 instead? That should collapse to a finite series which should work for all a and all x. It's only the non integer powers (and negative powers) we have to worry about. For example, (a+x)^(1/2).
 
Last edited:

MrAl

Well-Known Member
Most Helpful Member
Hi again PG,


Very good work there, and i see you are actually using your mind unlike a lot of other people that are in school :)

Just one question. You know that if we expand by a positive integer power we get a relatively short series that actually ends and does not require an infinite number of terms right? That means a power like 3, 4, 5, ... etc. will give us a closed series unlike with non integer powers. This makes sense because all we are doing is multiply (a+x) times itself a limited number of times. This also means we get an exact result rather than an approximation for any x and a. This also means it always converges.

If you'd like to see some other interesting series that dont have all these problems i can post one that is quite accurate and the domain isnt so limited.

BTW there are known mathematical tests for convergence which you may or may not be interested in.
 
Last edited:
Status
Not open for further replies.

EE World Online Articles

Loading
Top