how in the hell can I work out 2*x^{-1/4}

I cannot grasp the maths in my head no matter how many tutorials I watch, then I need to suss out how to get code to do it.

well nice challenge here for maths lovers.

the trick is to explain it in a way a devonrevenge can understand

I cannot grasp the maths in my head no matter how many tutorials I watch, then I need to suss out how to get code to do it.

well nice challenge here for maths lovers.

the trick is to explain it in a way a devonrevenge can understand

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Just think of it this way- anything raised to a negative exponent implies the reciprocal (assuming real numbers here), and anything raised to a fractional exponent implies that the numerator is the exponent that the variable is raised to, and the denominator the "root" of which the variable is taken of. In your case, it is 2 divided by the quartic root of x.

Any one know how to always find the centre of a line drawn between any two points on a Cartesian style grid (with the origin in the middle)

Are you familiar with search engines?

The link naraku9333 posted is the first link that pops up when you google "midpoint of a line segment".

[edit: Acknowledge the link posted by naraku.]

The link naraku9333 posted is the first link that pops up when you google "midpoint of a line segment".

[edit: Acknowledge the link posted by naraku.]

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oh yeah I have trouble with the tutorials, i though you would be more help, If you thnk about it you dont need c++.com you have google,

I was more thinking of the expression of the circle when you dont know the origin else its x1+x2/2 but my problem is deeper than that

I was more thinking of the expression of the circle when you dont know the origin else its x1+x2/2 but my problem is deeper than that

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If you thnk about it you dont need c++.com you have google, |

If you think about it, you're saying the time of everyone on this site is worth less than your own. Why should you google when your trusty cplusplus.com dupes can do it for you?

Some of the tutorials are extremely clear. Short comments posted on forum are not likely to be as informative. If the easy texts are too much to grasp, then cryptic verses are like maximum improbability drive: highly unlikely to hit the spot.

First exponents, then lines, now circles, and what has the origin to do with anything anyway? A wild goose chase?

First exponents, then lines, now circles, and what has the origin to do with anything anyway? A wild goose chase?

im struggling with my maths tutorials :/

@cire,

XD, but isnt that what every one who helps on beginner and general forum does?

TBF I plan on getting better so I can be of more help to people and im trying to get better by studying maths im finding the expression that is a circle reeally hard to follow, you know

x^{2}+y^{2}+2gx+2fy+C=O

I almost grasped it then lost it again

@cire,

trust worthy dupes |

TBF I plan on getting better so I can be of more help to people and im trying to get better by studying maths im finding the expression that is a circle reeally hard to follow, you know

x

I almost grasped it then lost it again

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Start from the distance to the origin function:

sqrt(x^2 + y^2) = r

A point (x,y) is part of a circumference of radius r [EDIT]centered at (0,0)[/EDIT] iff the above equation is true.

Now, say you want to the circumference to be shifted by 1 unit to the right (positive infinity of x axis). One's first inclination would be to modify the above equation to

sqrt((x+1)^2 + y^2) = r

Notice however that (r+1,0)*should* be in the circumference, but

sqrt((r+1+1)^2 + 0^2) = sqrt((r+1+1)^2) = r + 1 + 1 ≠ r

In reality, it should be

sqrt((x-1)^2 + y^2) = r

In fact, let (cx,cy) be the center, then

sqrt((x-cx)^2 + (y-cy)^2) = r

We can now start to simplify:

sqrt((x-cx)^2 + (y-cy)^2) = r

(x-cx)^2 + (y-cy)^2 = r^2

x^2 - 2 x cx + cx^2 + y^2 - 2 y cy + cy^2 - r^2 = 0

sqrt(x^2 + y^2) = r

A point (x,y) is part of a circumference of radius r [EDIT]centered at (0,0)[/EDIT] iff the above equation is true.

Now, say you want to the circumference to be shifted by 1 unit to the right (positive infinity of x axis). One's first inclination would be to modify the above equation to

sqrt((x+1)^2 + y^2) = r

Notice however that (r+1,0)

sqrt((r+1+1)^2 + 0^2) = sqrt((r+1+1)^2) = r + 1 + 1 ≠ r

In reality, it should be

sqrt((x-1)^2 + y^2) = r

In fact, let (cx,cy) be the center, then

sqrt((x-cx)^2 + (y-cy)^2) = r

We can now start to simplify:

sqrt((x-cx)^2 + (y-cy)^2) = r

(x-cx)^2 + (y-cy)^2 = r^2

x^2 - 2 x cx + cx^2 + y^2 - 2 y cy + cy^2 - r^2 = 0

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TBF I plan on getting better so I can be of more help to people and im trying to get better by studying maths im finding the expression that is a circle reeally hard to follow, you know x2+y2+2gx+2fy+C=O |

A circle is just defined as the set of points a certain distance (the radius) from some other point. So it makes sense that all points (x, y) a distance r from a point (a, b) could be represented by the equation

r = √(x-a)^2 + (y-b)^2, which is just the distance formula. If you square both sides and then expand the two terms you'll have something in the same form as the equation you wrote.

Edit: Akk, beaten to it.

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