| Andrew
Butt |
How does one go about expanding a bracket to the power of one half (ie. sqare-rooted)? NB the bracket contains four terms. Also, on a related note, how would you expand the following (I know how to expand the (x + y)^n): (z - (x + y)^n)^1/n |
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| Angelina
Lai |
Andrew, do you know Binomial Expansion/Series? It seems like you do, but in case you are not sure, here is an explanation from mathworld: http://mathworld.wolfram.com/BinomialSeries.html This works for non-integer values of n too. |
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| Chris
Tynan |
Angelina's link explains the Binomial series, but since your bracket contains four terms it is not a binomial so the series is not valid. As for expanding your second series, I'm not sure how to it, if you had it all in terms of one variable it is possible I think, by differentiating, but this is not the case. Chris |
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| Angelina
Lai |
Oh sorry, I totally blanked the rest of your post! What you can do is treating 2 terms as 1 at a time...but I realise that could be rather demanding algebraically... Looking at your expression, just wondered - is z meant to be complex? Because if so you can probably use De Moivres, which would make matter much simpler... |
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| Hauke
Worpel |
I'd make a substitution. Let a=(x+y)n Then you've got (z-a)1/n , which you can expand via the binomial expansion. When you've got your answer, change a back into (x+y)n and continue. |
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| Andre
Rzym |
Andrew, for your second question do you really want to expand powers of individual terms (x,y,z)? It's just that the most natural thing to do would be: (z-(x+y)n )1/n = z1/n * (1 -q)1/n = z1/n * (1 -(1/n).q/1! + (1/n)(1/n-1).q2 /2! - ...) where q = (x+y)n /z Chris, your approach will work fine if you use partial differentials (I'll show you if you're interested), but Andrew is this what you really want? Andre |
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| Marcos |
I'm interested Marcos |
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| Andre
Rzym |
Suppose you were asked to expand f(x)=(1+x)n as a power series. All you need to do is use a Taylor series expansion for f(x): f(x)=f(0)+f ' (0).x/1!+f ' ' (0).x2/2!+... where f ' denotes df/dx etc. [you will notice that if you substitute x=0; of differentiate then substitute x=0; or differentiate twice then substitute x=0 etc.; the series has the correct value for f(0) and all its derivatives]. Substituting: f(x)=1+n.x/1!+n(n-1).x2/2!+... Now suppose we have g(x,y)=(1+x+y)n The corresponding power series expansion is g(x,y)=[g(0,0)]+[gx . x+gy . y]+ [1.gx x.x2+ 2.gx y.x.y.+gy y.y2]/2! + [gx x x.x3+3. gx x y.x2.y+3.gy y x.x.y2+1.gy y y.y3]/3!+... where gx x y denotes ¶3 g/¶2 x ¶y etc. Notice the appearance of the binomial coefficients in the square brackets. You will notice again that with this construction, all the partial derivatives of g at x=y=0 are recovered exactly. Therefore in our example we have g(x,y)=1+n.[x+y]+n.(n-1).[1.x2+2.x.y+y2]/2!+n.(n-1).(n-2) [x3+3.x2.y+3.x.y2+1.y3]/3!+... This approach can be expanded further for more variables. Andre |
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| Marcos |
Thanks Marcos |
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| Chris
Tynan |
Thanks Andre Chris |
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| Andrew
Butt |
I think that my question may have been answered in there somewhere. I was investigating the differentiation of the 'circular graphs' as it were ie. y^2 + x^2 = k. Haven't got v. far, so i'll just look at something else, like y = e^x or y = sin x, which are easy. |
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| Andre
Rzym |
Andrew, your equation for a circle can be rearranged to y = ±(k-x2 )1/2 Do you know how to differentiate this (or were you trying to do something different)? In fact, if you are familiar with implicit differentiation, you can do it directly: y2 + x2 = k => 2y.dy/dx + 2x = 0 => dy/dx = -x/y Andre |
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| Andrew
Butt |
Aha. Thanks Andre. That was a fun GCSE investigation. (well actually we've finished the syllabus and started on A-level stuFF). |