Math Questions

[#1procrastinator]

Just thought I'd start a question thread as I don't really want to start a new topic every time I have a question...


First one, the question asks to find the fallacy in this argument:

sin ≣ sin(π - x)
Hence, f(sin x) = f(sin(π-x))
Let f(x) = x sin x
Then x sin x = (π - x)sin(π-x)
x = π - x
Hence π = 2x and since x is any value we choose, so is π

The answer says 'the fallacy is we cannot choose f(sin x) to be x sin x, this is not a function of sin x alone but of x and sin x'

But in the argument, he chose f(x) to be x sin x, what am I missing here? Is this an error? and what does he mean 'this is not a function of sin x alone but of x and sin x', he's referring to x sin x, right?

The line I think is incorrect is the 4th, where the x in front of the sin x is written as (π - x), but the topic is on functions

[dc302]

Because it's stated that f(sinx) = f(sin(n-x)), but in the next line, he does NOT use f(sinx) at all. He uses f(x).

for example, we could say that since f(sinx) = f(sin(n-x)), then if f(sinx) = sinx ^ 2, then sinx ^ 2 = sin(n-x) ^2. This is perfectly fine. However, the lines of reasoning make no relation between f(sinx) and f(x). It just chose some unrelated function of x alone, and let it be xsinx. Sorry for such a bad explanation


edit: actually I have a better argument, let me write it

We can think of this problem as a function that is constructed from multiplying two functions. Ie. let H(x) = f(x) * g(x) where f(x) = x and g(x) = sinx

So we know that g(x) = g(pi - x), but is it true that f(x) = f(pi - x)? ie, x = pi - x? Well of course not, so the statement is false.

Ie, we can't say H(pi - x) = H(x) because f(x) does not equal f(pi - x)

Hope this helps.

[Special At Specialist]

Following on from what dc302 said:
sin(x) = sin(pi - x) only when the function is sine.
If I were to change it:
f(x) = x*sin(x), then the function is no longer sine. The function is a variable coefficient multiplied by sine. Since the variable coefficient changes, so does the result.
So specifically:
f(x) = sin(x) and f(pi - x) = sin(pi - x), then f(x) = f(pi - x)
But if g(x) = x*sin(x) and g(pi - x) = (pi - x)*sin(pi - x), then g(x) does not equal g(pi - x).

[#1procrastinator]

Sorry for taking so -long to reply

Both posts make sense to me, but I'm confused as to what exactly f(sin x) means, I still can't see why he said 'we can't use f(sin x) to represent x sin x - what IS f(sin x)?  I'm used to seeing something like f(x) = x^2, where the value in the brackets is just the variable, and if it's f(x+2) = x^2, you replace x with (x+2).

So when I see f(sin x), I think of it as 'sining' the x-input before plugging it into the rule. If f(sin x) = sin x^2 as in dc302's first example, I'd think of it as it meaning the same thing as f(x) = sin(sin x)^2

So when he says f(sin x) = f(sin(π - x)), I interpret that as meaning if I plug in any number in, the two functions will give the same value so long as the rule is the same.

[dc302]

This is quite hard to explain, but perhaps best given through examples.

Here are correct examples:

f(xy) = (xy)^2 + 2xy
f(x+y) = 2x + 2y
f(sinx) = (sinx)^2 + 2sinx

Here are incorrect examples:

f(xy) = xy^2 + xy
f(x+y) = 2x + y
f(sinx) = x^2 + 2x

From these you should be able to tell why. Can you substitute the thing inside the function to make the RHS look like a 'normal' function?

For the first 3 examples, you can. Let me use the variable t:

f(t) = t^2 + 2t (letting t = xy)
f(t) = 2t (letting t = x+y)
f(t) = t^2 + 2t (letting t = sinx)

The last 3, you can't:

f(t) = ty + t (letting t=xy, and see how there is a leftover y)
f(t) = t + x (letting t = x+y, leftover x)
f(t) = (arcsint)^2 + 2arcsint (letting t=sinx) and here, arcsint is not a function.

[#1procrastinator]

So in the first 3 examples, would the 'naked function' (just f(x)) be the same thing as f(t)?

So f(xy) = (xy)^2 + 2xy is f(x) = x^2 + x if you didn't have the y in there?

EDIT: I think I kind of get what you're saying, but don't see how it ties in with f(sin x) = f(sin(pi-x) and then f(x) = x sin x

[kamil9876]

I think it's just giberish altogether.

Let f(x) = x sin x

If you define (a function from reals to reals) then it follows that

[dc302]

I'll try one more time but I'm not sure how to make this any easier to understand. Following from what kamil said,

let's revisit the original problem:

sin ≣ sin(π - x)
Hence, f(sin x) = f(sin(π-x))
Let f(x) = x sin x
Then x sin x = (π - x)sin(π-x)

Here we have 2 separate but correct/acceptable statements.

1. sin ≣ sin(π - x)
Hence, f(sin x) = f(sin(π-x))

2. Let f(x) = x sin x

However, these two statements are NOT related (for the purposes of this question). (1) makes a statement about f(sinx), there is NO mention of f(x). The two are not the same, clearly because x does not equal sinx. So what applies to f(sinx) does not necessarily apply to f(x).

[#1procrastinator]

This is from a physics text but it's more mathematical than physic..-sy

An object falls freely from height h. It is released at time zero and strikes the ground at time t. (a) When the object is at height 0.5h, is the time earlier than 0.5t, equal to 0.5t, or later than 0.5t?

I know you have to use h=0.5gt^2 but not really sure how to proceed. 0.5h = 0.5g(0.5t)^2? something like that? :S

[Special At Specialist]

More than 0.5t, because it's exponential.

The speed of its velocity downwards is increasing, so as you get further down, it takes less time to travel the same interval as an interval upwards.

So going from h to 0.5h is going to take more time than going from 0.5h to 0, since at 0.5h it will already have some initial velocity, whilst at h it has no initial velocity and must speed up.

Imagine it like a car starting at 0km/h and speeding up. The first 100m it travels will take ages, since it has to speed up, but after that it will be able to travel 100m a lot faster, since it is already at a fast speed.

[#1procrastinator]

^ Thanks, how would you find an explicit equation for it?

Also, how would you formally prove that product of the gradients of two perpendicular lines equals negative one? I've seen a few explanations but they didn't seem too formal (relied on visuals)

[dc302]

This is from a physics text but it's more mathematical than physic..-sy

An object falls freely from height h. It is released at time zero and strikes the ground at time t. (a) When the object is at height 0.5h, is the time earlier than 0.5t, equal to 0.5t, or later than 0.5t?

I know you have to use h=0.5gt^2 but not really sure how to proceed. 0.5h = 0.5g(0.5t)^2? something like that? :S

^ Thanks, how would you find an explicit equation for it?

Also, how would you formally prove that product of the gradients of two perpendicular lines equals negative one? I've seen a few explanations but they didn't seem too formal (relied on visuals)

Acceleration a at height h is given by: a= -g m/s^2 (acceleration due to gravity) and we shall write a = -g for clarity, for now.
Since a = dv/dt, where v is velocity, we can integrate both sides:
a = -g
dv/dt = -g
dv = -g.dt
v= -g.t + C
We know that the initial velocity is 0, which means at time t=0, velocity v=0, hence C=0

So v= -g.t

Since v = dx/dt, where x is position, we can again integrate both sides:

v = -g.t
dx/dt = -g.t
dx = -g.t.dt
x = -0.5g.t^2 + D, and since initial position is x=h at t=0, we get that D = h.

So x = -0.5g.t^2 + h

Now can you find the time taken for your question?

[Special At Specialist]

Also, how would you formally prove that product of the gradients of two perpendicular lines equals negative one? I've seen a few explanations but they didn't seem too formal (relied on visuals)

Well, we know that m = tan(θ) (if you want, I can derive that for you as well).
Now, if they are perpendicular, then:
Line 1 has a gradient of tan(θ)
Line 2 has a gradient of tan(pi/2 + θ)
Multiply them together:
tan(θ) * tan(pi/2 + θ)
= tan(θ) * tan(pi/2 - - θ)
= tan(θ) * cot(-θ)
= tan(θ) * -cot(θ)
= -1
QED

[kamil9876]

Also, how would you formally prove that product of the gradients of two perpendicular lines equals negative one? I've seen a few explanations but they didn't seem too formal (relied on visuals)

There is a quite easy visual proof based on similair triangles, I'm assuming you're reffering to one like that? If you want one which is 0% visual, use vectors (in fact that is how post 19th century math rolls, 0% visual(when working trying to prove things rigorously) so everything is replaced by vectors, real numbers etc. Is that the paradigm you are working in now?):

Let be the intersection point, we know that a line going through c can be represented as the set of all points of the form c+tw where w is a vector. Let's say and . So if the two lines are c+t(a,b) and c+t(a',b') then we have that the direction vectors are perpendicular hence their dot product is 0:

[#1procrastinator]

@dc302: when i said 'explicit equation', i meant how h relates to t, so if it falls 0.5h then what fraction of time would that be (hope i'm making sense)

@Special At Specialist: how did you go from tan(pi/2 - - θ) to cot(-θ) (i need to brush up on my trigonometry)

@kamil9876: thanks, i should be able to better appreciate your proof in a few days. yeah, I prefer to be able to work things out without visuals whenever possible. what is it called when you work things out like this? is it analytically?

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how would you solve this inequality?



I've tried




then setting one binomial positive and the other negative and vice-versa won't give the correct answer
i'm getting and

i know i've made a silly mistake somewhere, but it's invisible to me at the moment