ahh, i got it now. thanks.

Be a little careful though - although there are two solutions to the quadratic, only the positive root corresponds to a solution to the original problem. This is because if x is negative, log(x) doesn't make any sense.

What was the longest homework assignment you've ever turned in? Today I handed in a 26-page assignment for my separations class, which I spent like 15-20 hours working on. Granted about half of it was graphs like this:



But drawing those "steps" for each stage accurately is kind of a bitch.

I am trying to figure this out and it is bringing up some questions.
x^3=x*x*x
x^2=x*x
x^1=x*1=x
(First question is, what is the justification for assuming we are multiplying by 1 here? It seems axiomatic how is it justified)
x^1=*1 (again if it's just assumed that 1 is what we are multiplying by we get 1*1=1)

x^5/x=x^4
x^1/x=^1

again I don't quite understand why it's 1 we are raising to the ^1?
Why is it not 0^1

You can always multiply by 1. Your expression would be the same with or without it. Hell, you can multiply by 1 on the other expressions too.

0^1=1 which is what you're looking for?

It's very clear you don't know the definition of integer powers. It's impossible to prove anything about an object you don't know it's defined.

Positive integer powers of a base can be defined as a "short notation" for multiplying the base against itself a^n = a.a.a ..... a (n times).

Now, do you see why a^2=a.a and why a^1=a ?. It's pretty much by definition.

I'll get you started on proving the properties of powers. What happens if you multiply (a^n).(a^m) ?. Well, just expand the expression based on the definition:

(a^n).(a^m) = (a.a.a ... ).(a.a.a.a.a ...)
= a.a.a.a ..... a

Because you have n elements a in the first parenthesis and you have m elements in the second, then how many "a"s do you have in total multiplying together?

You have (m + n) total "a"s multiplying together, so how do you write this number using power notation?. Well, by definition, "a" multiplied by itself (n + m) times is a^(n + m).

So we have just proved using only the definition of power notation that:

(a^n).(a^m) = a^(n + m)

Try to do division of powers (same base) where the power in the numerator (n) is larger that the denominator (m):

Meaning, why is: (a^n)/(a^m) = a^(n - m) ?

Then think about what the result should be if n is equal to m (dont think of powers, just think of the fraction).

I'm trying to compute 3 Bonferroni confidence intervals for 3 groups. Do I use the error I used to calculate the f statistic(MSE) for all 3 of the groups in all 3 intervals or only the 2 I'm comparing at the time

x^1 isn't really x*1. When you have x^n you write down x n times and write down (n-1) multiplication symbols. so x^1 is just x with 0 multiplication symbols.

This question is driving me nuts. Solve for x:



Isn't this true?



So... 1 = 3???

Hint: Note, there are 2 possible solutions, A and B, where A < B. A is invalid and B is valid.

No idea how to approach this! Thanks.

The log of a square root isn't equal to the square root of the log.

Log(a^b) = b . Log (a)

In your case b= 1/2

Most functions don't have the property that f(g(a)) = g(f(a)), but when in doubt try a few values (e is a good one in your case)

Thanks, Chasqui. Still getting stuck here, but let me take it a few steps further.

Solve for x:







So far so good?

Noooow, can I do this?



Doesn't seem right, because then:



Clearly I'm confused about how to properly distribute exponents in log functions.

When you square Ln(b) you get (Ln(b))^2. Another instance of "the square of a log is NOT the log of the square.

Hint: look at the fraction and don't try to solve for x, solve for Ln(x) as your unknown first. You'll have to be familiar with the properties of powers for this.

After step one (ln(sqrt(x))=0.5ln(x)), you can sub y=ln(x) and solve for y. Then solve for x.

Or you can do what you did and exponentiate both sides.

edit: oh, you just said that right above me. Sorry.

Thanks, Chasqui and Wyman.

So, finally we have:

Solve for x:





Let ln x = y (this made it really easy for me, and we've used this trick before in class a few times, but it slipped my mind today...)









y = 0 or 1/2.25

ln x = 0 or 1/2.25

x = e^0 or e^(1/2.25)

x = e^0 is invalid since the fraction would be undefined.

Thanks!!

Looks good to me. But expressions like 1/2.25 make my eyes hurt; please just write it as 4/9.

You may want to claim in #2540 that x=1 is not a solution but more appropriate way i think is to say that by setting ln(x)=y since ln(x) has to be >= 0 for the original expression to be in real numbers this means x>=1 is a starting condition and furthermore establish that x->1 limit is not satisfying the expression (its left part 2 as x->1+ and right 3) so 1 cannot be a solution of such continuous expression as the limit x->1+ of that expression is not 3. I think thats a more appropriate way to eliminate 1 since on occasion you may get cases that the expression is not defined at some "solution" but the limit exists and satisfies the equation. I think such cases are interesting enough to deserve a separation even if in the strict sense the expression are not defined at the root. I wonder however if such cases are in fact seen as solutions (provided that it can be shown such cases exist without ending in something inconsistent) or some theorem not immediately evident tome easily excludes them from existing as solutions of the resulting process.

I don't think such (hypothetical) cases should count as a solution.

The situation you're describing seems most likely to arise, IMO, when you're asked to find the solution to some equation f(x) = c on some open domain U, and you discover that there exists a point x_0 on the boundary of U such that lim(x \to x_0) f(x) = c. I would still never call x_0 a solution, though, because it doesn't lie inside U.

having a bunch of trouble with this problem, would appreciate some input.



i think i can get pointwise convergence but showing the 0<P_n<P_n+1 < x and showing that the P_n's are a monotonic increasing sequence with upper bound x, and refine it a little to negative values.

however i'm having alot of trouble showing that || |x| - P_n ||(infinity) converges to 0 (convergence in the sup norm)

Let y denote P_n(x) [I don't want to open LaTeX]

Consider |x - P_{n+1}(x)| = (x-y)(1 - (x+y)/2 ) < (x-y)(1-y).

But consider this case; (re #2450 subthread)

Solve -1/2*(x - 1)^2*x^2/(x - e^(x - 1))=1

You cant deny that x->1 is at least an asymptotic type of solution that is also a solution of the process that solves this that was used in the problem above. And it has also another 2 clean solutions near -0.77 and 8.74 in the strict sense of the word (the domain in which the expression is defined without singularities).

Otherwise one must start by demanding the denominator is never 0 (or in the original problem that x>1 to define the ratio of the log-functions) and proceed to solve it eliminating the x=1 problem but in the process losing something essential as in my example here (but not the original homework problem) ie that the limit satisfies the expression and so the value of the starting function at x=1 may not exist but the limit does, making the function less pathological at that point than some infinite limit.

I understand sometimes i talk about these things from the point of view of a physicist for whom there is no such thing as number 1 in many real life problems when you are looking for solution to physical problems where a particular math study is only an approximation of reality and for me clearly 1 would be a solution because something like 0.999+-0.002 easily takes that expression within a range of the required ratio of 1 i demanded in trying to solve it.

In any case to be rigorous enough one needs to do in my opinion these 2 things in problems of such nature to avoid missing a thorough understanding of the investigation;

1) First determine the domain where the original function in the equation that needs to be solved is defined. Detect all pathological points and eliminate them in the methodology that follows. Find the solutions in the typical sense if possible.

2) Then come back to the original expression and see how it behaves in the limit of the pathological points. If the limit satisfies the equation then a note must be made to produce a much richer answer in my opinion.

Hi, can someone please tell me what I did wrong with this Statics problem because I thought I did everything correctly.

Bellow I have the question, followed by the diagram, followed by the work I did to solve the problem.

Ramp ABCD is supported by cables at the corners C and D. The tension in each cable is 810N. Determine the moment about location A due to the force exerted by cable D and the force vector TsubDE. (Note: Although difficult to see in the diagram, point B is located 1m above the base).




Note: I'm not sure what the k component of the moment is supposed to be, but I know my professor said the j component is "0", which makes no sense to me.

Actually when everything is continuous, monotonic pointwise convergence on a compact => uniform convergence
Dini's theorem

^^ thanks helped a lot

Is there some sort of rounding issue? You got pretty closet to 0.