solving equations but they get increasingly more impossible?

We will solve 4 impossible-looking equations, sqrt(x)+sqrt(-x)=2, ln(x)+ln(-x)=0, e^x+e^(-x)=0, and sin(x)+sin(-x)=2. From verifying the domains of the functions to finding real solutions, we go step-by-step through each equation. And just when you think you've got it figured out, we hit you with one more equation that's sure to blow your mind! Get ready to flex your mental muscles and solve some of the trickiest equations out there. #math #equation #blackpenredpen

0:00 sqrt(x)+sqrt(-x)=2
3:27 ln(x)+ln(-x)=0
6:10 e^x+e^(-x)=0
10:29 sin(x)+sin(-x)=2

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#maths #equations #blackpenredpen

[oreocookiedough]

Putting these no solution questions as a pre-cal bonus question saying "graph the function" would be humorous when the people who don't know what to do leave it blank and then you mark it correct.

[ecg]

The last one has no solution at all because sin(-x) is equal to -sin x so the equation sin x+sin(-x)=2 is the same as saying sin (x)-sin(x)=2 now we can cancle the two sines and we get the equation 0=2 and it has no solution.

[ffggddss]

Here's an alternate take for the exponential equation.
That parabola-like curve is a catenary; and eˣ + e⁻ˣ = 2cosh(x); twice the hyperbolic cosine. But from Euler's formula, (circular) cosine can be written
cos(x) = ½(e ͥˣ + e⁻ ͥˣ) = cosh(ix); likewise, because cos and cosh are even functions,
cosh(x) = cos(–ix) = cos(ix)

So eˣ + e⁻ˣ = 2cosh(x) = 0, means that
2cos(ix) = 0 = 2cos(–ix)
But we know where cosine is 0:
–ix = (n+½)π ; x = (n+½)iπ

And that's another way to solve this one.

Fred

PS. Great idea, this set of problems!

[Paul-]

For #3, it’s interesting that if you substitute e^i(pi) = -1 too soon or too late, you get stuck with a tautology where x = anything.

[mrgreenskypiano]

This is honestly one of my favorite videos of yours - it’s very clear and concise but still enough content to fill a 10-minute video! All of these sections are different but have the same theme so it still feels like one video - my rating is (pi^2 + 1)/10

[maxwellcody]

It looks like sin(x)+sin(-x) = 2 has no solutions since sin(x) is an odd function, even with using Euler's formula it leads to a contradiction of 0=2.

[justintroyka]

If you, like me, are bothered by the fact that the process in part 1 doesn't yield both solutions, then read on. The reason is because a number has two square roots, and complex numbers don't have a preferred one of the two like positive real numbers do - you can't just take the "positive" square root because most complex numbers don't have a purely real, positive square root. More specifically, the problem is in the step √[-x] = i√x. Check this out:
√[-(-x)] = i√[-x] = i*i*√x = -√x.
So √x = -√x, which is absurd. The way to resolve this is that √[-x] may be either i√x or -i√x, depending on which square root of a number is being chosen.

[arostheautistic]

**gets paper ready for last question**
**realises sine is a strictly odd function**
nvm

[zactron]

2:45 the second solution can also be found by observing that taking out a factor of i and taking out a factor of -i are both equally valid starting points ((-i)^2 = i^2 = -1). So in reality, you needed to take out a factor of +-i rather than just +i

[en]

1:49, that's when you lose the other solution, sqrt(-1) is i OR -i, so you should get two equations

[akashsriram]

For the e based expression, I used cosine def in terms of e so we get (e^x + e^-x)/2 = 0 = cos (-ix)
Take the inv cos from both sides to get pi/2 = -ix and solve for x to get -i*pi/2

[logicxd]

the last one has no real nor complex solution, because if we use the complex definition of sinx, we have = 2, then e^(ix) - e^(-ix) + e^(-ix) - e^(ix) = 4i, which LHS cancels out to 0, we have 0 = 4, no solution

[IvoCampi]

The solution of e^x + e(-x) = 0 is straightforward if thinking the two terms as two vectors in the complex plane, having the same module. For simmetry, their phases should be pi/2 and -pi/2

[luvHugeBressts]

İ can do that! İn the last question, it have no solution. Because if we try to do that, we get 0=2 and that was an issue.

[AntonFediukov]

8:45 can you use the Lambert W function if you multiply both sides by x? That yields an extraneous solution x=0 for one of the branches but I was wondering if this method is viable here if you know what you’re doing

[miscccc]

Petition to make I dont like to be on the bottom, I like to be on the top merch.

[TeamBuster]

Mathematician: "Infinity is not a number, therefore this equation has no solution"
Me, a physicist: "I have no such weakness"

[antonyzhilin]

I have no idea where to start with the last problem, it's so odd tbh

[xavierwainwright]

sin(x)+sin(-x)=2 <=> sin(x)-sin(x)=2 <=> 0=2. There are no solutions.

[justushinkelmann]

Holding the poceball gives him the ultimate mathematical power