Nah thats not it, the program always tells you if you put extra brackets (I’ve been through that error before lol)
I just don’t know what it wants.
@HoneyBBQGrundle
Is there a way to find summation formulas by hand without having to work backwards. Me and a buddy accidentally found the formula for a triangular number when we tried to replicate finding the summation of 1 to 100. And we were able to successfully apply this case to all non-negative numbers.
However, I thought when doing math, “working backwards” could not operate as a general proof.
I don’t know how fancy it is, but maybe it wants (n+1)^2/n^2 instead of [the equivalent] (n^4+n^2)/n^4 ?
thx, ill check
I’m not sure what you mean by working backwards (needing the formula in the first place to apply induction maybe?). You can do anything you want in math to discover the idea behind a proof, and then later prove it rigorously by whatever method you prefer. That’s how math is actually done in practice (a not so good example is how in epsilon-delta proofs one generally finds the value of epsilon last and then reverses the steps). I love the sums of first, second, third powers because there are so many ways to prove them, even if not all of them are fully “rigorous”–and some of the best ones aren’t. I actually just learned of another way a few days ago.
If anyone can show me the steps to solving this problem, it would be greatly appreciated.
I just can’t get to the right answer. It’s a difference quotient
[1(x+h)^2-1/x^2]/h
I’m assuming you mean [(x+h)^-2 - x^-2]/h. Write it out as the difference of the two rationals over h. Divide each by h: for example, the first denominator should be h(x+h)^2. Get each to a common denominator and make it a single fraction. Simplify the numerator (you should get -(2x +h) in the numerator).
For the derivative, just take the limit as h approaches 0. It’s a really simple one, no L’Hopital’s rule or anything needed.
I’m re-learning math on Khan Academy. Currently taking Algebra. I’ll update my progress.
:tup:
Khan Academy is probably a bad way of learning from what I’ve seen…but if it suits your needs, the I think that’s fine.
I figure one of you guys might know if this can be done.
I have a calculation that we used to use a lookup table to estimate one variable. A few years ago, I found a curve/surface fitting tool online (source files still exist - ZunZun) that let me use that lookup that lookup table as a series of points to create a proper formula that was overall superior than the lookup table (as the lookup table was jank as hell and occasionally values fell out of range to properly estimate). Sadly that site is apparently depreciated and I can’t recreate the surface fitting, but I still have the formula that we use to this day.
The problem I’ve had has been the fact that this formula pulls some crazy values. One variable is affected by the power of 4/37 (0.108), and another variable is actually a ratio of two variables (x/y) to the power of -2/3 (-0.667).
My question to you eggheads is this: can (x/y)^(-2/3) be simplified in any way? I know the whole z^(-2/3) is the same as 1/(z^2/3), which would be the same as 1/(cuberoot(z^2)), but going from there considering that z is actually x/y, I wasn’t sure if this could be factored out further, or is that crazy negative fractional exponent of a ratio just going to be as convoluted as that and remain this way?
I’m not sure the context in which you want the simplification, but you can take the nth root in the complex plane. Thats only thing that comes mind
Mostly I was just making sure it didn’t just simplify into (y/x)^(2/3) or something like that. Also:
https://media.riffsy.com/images/086549810e170bcbba237e81ec5901d1/raw
So…
1/0=∞