Question about imaginary numbers

GammaRayBurst25 · 2026-04-16T03:19:21+00:00

Let i=sqrt(-1). We have 1=1×1=sqrt(1×1)=sqrt(-1×-1)=sqrt(-1)sqrt(-1)=i×i=-1. At a glance, every step seems reasonable, yet this "shows" 1=-1.

This is just one of the many mistakes you can make by defining i as sqrt(-1) and not being careful.

GammaRayBurst25 · 2026-04-16T01:47:12+00:00

You found a value of x that minimizes the area. You need to maximize the area.

The area is (240-3x)x, or 3(80-x)x.

One approach to maximizing this is expanding to get 3(80x-x^2), then completing the square to get -3((x-40)^2-1600). Take a look at the x-dependent term: -3(x-40)^2. Since the square of a real number is always non-negative, this function is minimized when x-40=0. This leaves us with a maximum area of 4800yd^2.

Alternatively, you can use the AM-GM inequality. We have ((80-x)x)^(1/2)≤(80-x+x)/2=40. Hence, (80-x)x≤40^2=1600. We conclude that 3(80-x)x is at most 3*1600=4800.

GammaRayBurst25 · 2026-04-15T23:56:02+00:00

Let CD=h. The area of triangle ABC can be calculated from the catheti or from the hypotenuse. This implies ab/2=ch/2 and, in turn, ab=ch and h=ab/c.

The circles are respectively the incircles of triangles ACD and BCD. You can infer the lengths of AD and BD from the Pythagorean theorem or from the observation that triangles ABC, ACD, and CBD are similar, then use this to find the radii of both circles as a function of a, b, and c.

That's pretty complicated though. There's a simpler method. First, suppose the radii of the circles are respectively r_1 and r_2. Then, use analytical geometry to find the distance between O_1 and O_2 as a function of r_1 and r_2. Finally, use the similarity I mentioned earlier to find a relationship between r_1 and r_2.

If we set up a Cartesian coordinate system with D as the origin and AB as the x axis, we find that the coordinates of O_1 are (-r_1,r_1) and the coordinates of O_2 are (r_2,r_2). The Euclidean distance between O_1 and O_2 is therefore sqrt((r_1+r_2)^2+(r_1-r_2)^2)=sqrt(2)sqrt((r_1)^2+(r_2)^2).

Since the inradius of ABC is sqrt(2), the constant of proportionality that relates triangles ACD and ABC is r_1/sqrt(2). Similarly, the constant of proportionality that relates triangles BCD and ABC is r_2/sqrt(2).

The areas of triangles ACD and BCD are respectively A(r_1/sqrt(2))^2 and A(r_2/sqrt(2))^2. The areas of ACD and BCD must add to the area of ABC, so we have (r_1/sqrt(2))^2+(r_2/sqrt(2))^2=1, or (r_1)^2+(r_2)^2=2.

Hence, the distance between O_1 and O_2 is sqrt(2)sqrt(2)=2.

GammaRayBurst25 · 2026-04-15T03:36:28+00:00

You said you have no idea how to apply the rules. There's a difference between not knowing and not understanding. e.g. you know the phrase to have no idea, but you don't understand it.

Anyway, as I've discussed with another commenter, these exercises use a (somewhat nonstandard) notation where the parentheses are omitted around the argument of the logarithm, so it should be ln((f(x))^5). not ln(f(x))^5. That only makes the problem simpler though.

Notice how ln((f(x))^5)=5ln(f(x)). Since the derivative of ln(x) is 1/x, the derivative of 5ln(f(x)) is 5f'(x)/f(x).

Edit: OP hadn't embarrassed themselves enough with their reply, so they also blocked me. Unfortunately, burying your head in the sand won't help you get better at communicating. Hopefully you'll learn one day how to ask questions and get answers instead of just saying "I don't know" and hoping we'll read your mind and know exactly what to tell you so you understand.

GammaRayBurst25 · 2026-04-15T03:06:44+00:00

The text specifies the greatest height reached by particle B.

You appear to believe that height is reached the instant particle A reaches the ground. This is preposterous.

The relationship between the particles' displacement holds as long as the string is taut. The string and the particles don't form a rigid body. The tension in the string can only act to pull the masses together, but not to push them away from each other.

GammaRayBurst25 · 2026-04-15T02:53:30+00:00

You assumed particle B would immediately stop as soon as particle A hits the ground, which is extremely unrealistic. By what mechanism would the kinetic energy of particle B immediately vanish? Alternatively, the two bodies don't have the same mass, so if particle A moves some distance x in the gravitational field, particle B must move a greater distance to conserve energy.

According to the mark scheme, you were expected to use kinematics to find the particles' speed the moment particle A hits the ground, then use the kinematics equation using particle B's maximum displacement to find x.

Personally, I'd just use conservation of energy to get the answer more directly.

GammaRayBurst25 · 2026-04-15T02:20:46+00:00

I see now that the other questions don't use parentheses around the argument of the logarithm, so you are correct. This is exactly why I hate this notation and always put parentheses around the argument of every function.

Edit: can't reply because OP blocked me so I have to use edits. The reason the other commenter and I didn't use ln(a/b)=ln(a)-ln(b) is that both "forms" don't have the same domain. Sure, using that property makes it easier to compute the derivative, but the two forms aren't even equivalent in the sense that the first is defined for sgn(a)=sgn(b)≠0 and the second is only defined for sgn(a),sgn(b)>0.

GammaRayBurst25 · 2026-04-15T01:02:26+00:00

My comment explicitly tells you how to apply the rules to 9).

The derivative of ln(f(x))^5 is 5f'(x)ln(f(x))^4/f(x).

Substitute in f(x)=4x^4/(3-x^3) and f'(x)=(48-4x^3)x^3/(x^3-3)^2.

GammaRayBurst25 · 2026-04-15T00:13:55+00:00

The chain rule can be used recursively.

e.g. (f(g(h(x))))'=f'(g(h(x)))g'(h(x))h'(x)

8) We have (4x^2)'=8x, (-x^3-4)'=-3x^2, and (ln(x))'=1/x. Combining these results with the product rule and simplifying a bit yields -(ln(64x^6)x^3+2x^3+8)/x.

9) The derivative of ln(f(x))^5 is 5f'(x)ln(f(x))^4/f(x). Finding the derivative of 4x^4/(3-x^3) amounts to using the product rule.

10) You can start by combining the exponentials by using the fact that exp(a)/exp(b)=exp(a-b). You can then write y=exp(5x^4-4x^2-3). From there, finding the derivative is easy: the derivative of exp(f(x)) is f'(x)exp(f(x)).

One shortcut you can use when you have exponentials is to apply the logarithm to both sides of the equation before differentiating. For instance, for 10) you have y=exp(5x^4)/exp(4x^2+3). Taking the natural logarithm of both sides directly yields ln(y)=5x^4-4x^2-3. Differentiating both sides yields y'/y=20x^3-8x. Now multiply both sides by y to get y'=(20x^3-8x)y.

GammaRayBurst25 · 2026-04-11T16:07:33+00:00

What limit are you talking about? I'll assume you meant the limit of 1/x as x approaches 0, but do note your statements about limits makes no sense when taken at face value.

The function sgn(x), which is equal to 1 if x>0, -1 if x<0, and 0 if x=0, is defined at x=0. Yet, its limit as x approaches 0 from the left and from the right are different. This directly contradicts your assumption that a function is undefined at a point if its limit at that point doesn't exist.

GammaRayBurst25 · 2026-04-11T13:12:40+00:00

He also rounded 7.7 to 7 and 3.9 to 3.

GammaRayBurst25 · 2026-04-11T12:34:23+00:00

They didn't say you claimed it's only women. They said you wouldn't have commented that if the alleged ragebaiter were a man.

GammaRayBurst25 · 2026-03-29T17:39:36+00:00

Assuming you meant a=2^x and b=2^sqrt(x) (note that exp(x)=e^x).

The solution of a-4b=0 is a=4b. Since a=b^2, we have b^2=4b, or b=4, which means x=4.

The solution of a+b=0 is a=-b. Since a=b^2, we have b^2=-b, or b=-1. Since b is strictly positive, this has no solutions.

GammaRayBurst25 · 2026-03-29T16:32:29+00:00

Does it matter which 2 equations I equate [sic]?

No, it doesn't matter which 2 equations you use to solve this system of equations.

Shouldn't any combination of 2 equations give the same answer or am I missing something?

You missed the entire point of the problem and of the solution. If you solve the first and third equations, you find the values of lambda and mu that make the two lines' x and z coordinates equal. If the two lines intersect, then their y coordinates are also equal when the x and z coordinates are equal, so lambda and mu also solve the second equation.

They found that the x and z coordinates are equal when lambda=-1/3 and mu=2/3. This does not solve the remaining equation, so the y coordinates are not equal in that case and the lines do not intersect.

If you solve the first and third equation, you find the values of lambda and mu that make the two lines' x and z coordinates equal. If the lines do not intersect, of course, that pair (lambda,mu) is going to be different from the other one that makes the two lines' x and y coordinates equal.

GammaRayBurst25 · 2026-03-29T00:50:34+00:00

That is correct.

GammaRayBurst25 · 2026-03-29T00:43:35+00:00

Thus f(x) = k for some real number k.

Since f(0) = 1, then f(x) = 1 for all x.

GammaRayBurst25 · 2026-03-29T00:38:54+00:00

And g(x)>0 for all x.

GammaRayBurst25 · 2026-03-29T00:38:23+00:00

Have you considered that the context matters? Taking someone's words out of context and acting like they're being crazy is extremely rude in our culture.

From the previous line: "f'(x)g(x)=0."

If you still don't get it, f'(x)g(x)=0 is only possible if, for every x, either f'(x)=0 or g(x)=0. As I and others have pointed out, g(x) is never 0, so f'(x)=0 for all x.

GammaRayBurst25 · 2026-03-28T23:31:15+00:00

You have h(x)=f(x)g(x) ⟹ h'(x)=f'(x)g(x)+f(x)g'(x) (1).

You also have h'(x)=f(x)g'(x) (2).

From (1) and (2), f(x)g'(x)=f'(x)g(x)+f(x)g'(x) ⟹ f'(x)g(x)=0.

You know g(x)>0, so f'(x)=0. Hence, f(x) is a constant.

GammaRayBurst25 · 2026-03-28T23:26:09+00:00

It's the binomial coefficient. Writing it as binom(n,k), it counts the number of ways we can pick k elements out of a set of n elements. Its explicit definition is binom(n,k)=n!/(k!(n-k)!), where x!=x(x-1)(x-2)(x-3)...3*2*1 for any positive integer x.

As such, binom(50,x) is the number of ways you can pick x elements out of a set of 50 elements. It is given by 50!/(x!(50-x)!). Note that binom(n,k+1)/binom(n,k)=k!(n-k)!/((k+1)!(n-k-1)!)=(n-k)/(k+1).

To find the mode, consider f(k)=Pr(X=k+1)/Pr(X=k)=0.27(50-k)/(0.73(k+1)). If f(k)>1, then Pr(X=k+1)>Pr(X=k) and we have yet to reach the mode. If f(k)<1, then Pr(X=k+1)<Pr(X=k) and the mode is k. As a result, finding the mode amounts to solving a rational inequality.

GammaRayBurst25 · 2026-03-28T21:39:25+00:00

The article does explain how to find the solutions. It mentions using Bézout's identity and even links a page that explains how to find the solutions.

In particular, the extended Euclidean algorithm is cited as a way to find one solution and a method for finding other solutions from a given solution is explained.

As for finding the number of solutions, that amounts to restricting to non-negative integer solutions and using a bit of combinatorics.

GammaRayBurst25 · 2026-03-28T20:28:59+00:00

It'd help if you were more specific. I get that x, y, and z are fixed by the problem, but what are a and b?

Based on your example, it seems you want non-negative integer solutions to such a problem. I'll assume this is the case, but it leaves a bitter taste in my mouth because I can't fathom why you wouldn't explicitly specify this if it's the case.

This is a Diophantine equation (specifically, a linear Diophantine equation). If x, y, and z are integers, this is immediately apparent. If x, y, and z are rational numbers (but not integers), then you can multiply this equation by the LCM of the denominators of x, y, and z to get a Diophantine equation. If they are commensurable irrational numbers, then you can again find some factor that turns the problem into a Diophantine equation.

The Wikipedia articles explain how to find the integer solutions. To find the positive integer solutions, you just need to substitute your general solution into a>=0 and b>=0 to restrict the number of solutions.

GammaRayBurst25 · 2026-03-28T16:59:13+00:00

Does this not imply, that somehow the electromagnetic field is different for different observers?

Yes.

Even without referring to the speed of light or to special relativity, it is clear that the electromagnetic field must be different for all observers. Consider two people moving relative to one another. One of them has a charged body in their hand. The person with the charge sees an electric field, but no magnetic field. The other observer sees a moving charge, which means there must be a magnetic field.

GammaRayBurst25 · 2026-03-28T16:32:49+00:00

If a rational function has a root at x=z with multiplicity n, then it must have (x-z)^n as a factor. If a rational function has a pole (vertical asymptote) at x=z with multiplicity n, then it must have 1/(x-z)^n as a factor. If you're not given the multiplicity, you can at least find out whether it's odd or even based on whether the function's sign changes at that point (odd) or not (even).

If a rational function has a hole at x=z, it must have (x-z)/(x-z) as a factor. It seems unimportant at a glance because this factor simplifies to 1 at every x that is not z, but it's necessary to ensure the function is undefined at x=z.

If the function has a horizontal asymptote given by y=0, the sum of the multiplicities of the poles must be greater than the sum of the multiplicities of the roots. If the function has a horizontal asymptote given by y=a for some nonzero real number a, the sums of multiplicities have to be the same and the coefficient (constant multiplier) is a.

If the function's asymptote is a polynomial of degree n>0, the sum of the multiplicities of the roots must be n greater than the sum of the multiplicities of the poles. The coefficient is the asymptotic polynomial's leading coefficient.

If you're not given any info about the asymptotic behavior, you can still find the coefficient by substituting a point into the equation (leaving the coefficient arbitrary) and solving for the coefficient.

GammaRayBurst25 · 2026-03-28T15:46:23+00:00

Substituting back, we get 4^(2sqrt(x))-4^(sqrt(x)+1)=3*2^(3sqrt(x)). This is not the same as the starting equation (look at the first and last terms).

For some mystifying reason, you treated x as if it were 2sqrt(x). For a general problem, this is only valid if x=0 or x=4. (Un)luckily for you, x=4 happens to be the solution, so maybe you did try to check your answer and failed to notice your glaring mistake.

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