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I jsut wanted to ask:

Root of (3^x + x)

Is this (3^x + x)^1/2

And would this lead to 3^0.5x + x^0.5?

Thanks.
Reply 1
Original post by Tynos
I jsut wanted to ask:

Root of (3^x + x)

Is this (3^x + x)^1/2

And would this lead to 3^0.5x + x^0.5?

Thanks.

Yes

I think yes
Reply 2
Original post by Tynos
I jsut wanted to ask:

Root of (3^x + x)

Is this (3^x + x)^1/2

And would this lead to 3^0.5x + x^0.5?

Thanks.


Original post by krisshP
Yes

I think yes

No, no, no! :eek: No it most definitely won't! :eek:

The square* root is equivalent to taking something to the power of a half, so the first part is correct. In the second part you have assumed that it has a similar property to multiplication, but unfortunately no such simple property exists.

If this is the sort of thing you are learning at the moment then it is unlikely that you will have to "expand" the bracket, but if you are interested, look the the "Binomial Expansion". There is a version for non-integer powers that is derived from "Maclauren's Series" (stretching off to infinity). In the case of the expression you have, this will require an initial factorisation.

Hope this helps! Even though it makes things horrendously more complicated (infinite convergent series instead of a two-term expression), mathematics is about truth not convenience :smile:
Reply 3
Ah, I was reading root as in finding the root of an equations, not the square root.

As for finding the root (as in root of equation), I am not sure that you can do it analytically unless there is a series solution of some kind which I do not know how to do. (If someone knows how to do this then I would be interested).

If asked to find the root, then you could use Newton Raphson as the function is differentiable and well behaved.
(edited 10 years ago)
Reply 4
Original post by Jkn
No, no, no! :eek: No it most definitely won't! :eek:

The square* root is equivalent to taking something to the power of a half, so the first part is correct. In the second part you have assumed that it has a similar property to multiplication, but unfortunately no such simple property exists.

If this is the sort of thing you are learning at the moment then it is unlikely that you will have to "expand" the bracket, but if you are interested, look the the "Binomial Expansion". There is a version for non-integer powers that is derived from "Maclauren's Series" (stretching off to infinity). In the case of the expression you have, this will require an initial factorisation.

Hope this helps! Even though it makes things horrendously more complicated (infinite convergent series instead of a two-term expression), mathematics is about truth not convenience :smile:


Ahh ok, thanks bro.

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