Let's just talk about unit quaternions.
I know that $$\left(\cos{\frac{\theta}{2}}+v\sin{\frac{\theta}{2}}\right)\cdot p \cdot \left(\cos{\frac{\theta}{2}}-v\sin{\frac{\theta}{2}}\right)$$
where ##p## and ##v## are purely imaginary quaternions, gives another purely imaginary quaternion which...
What multiplication basically achieves in these situations counting the units. Draw a 1cm x 1cm square and call its size 1 unit area. You can't measure how big the unit is. You can only take a look at it and see how big it is. However, you can measure the size of other surfaces in terms of the...
I personally don't think saying that something can't be known and true at the same time is the same saying something is an unknowable truth. The sentence 'p is an unknown truth' is just a clever sentence which is no longer true when p is known. This sentence is not really an unknowable truth...
So you're saying that there can't be a finite number of truths. But I don't think that makes much difference. In such a universe, infinite number of truths written on different pieces of paper are floating in the universe. All papers are numbered. The person has a device in which they have to...
https://en.wikipedia.org/wiki/Fitch's_paradox_of_knowability It begins by assuming that 'All truths are knowable' and then logically proves that that assumption means 'All truths are already known.' The proof is like this:
Suppose p is a sentence that is an unknown truth; that is, the sentence...
Honest trailers named Force awakens 'Star wars: A familiar hope' lol. It was almost a remake but it sure was entertaining. I think the next two movies should be original. I'll lose interest if they remake again.
I was reading this wikipedia article: https://en.wikipedia.org/wiki/Ant_on_a_rubber_rope
In the analytic solution section, it says that, at time ##t##, a point at ##x=X##, i.e. initially at a distance ##X## from the starting fixed point on a uniformly expanding rope moves with a speed...
I have a great explanation of parameters. I read it on a website:
For every degree above 70, our convenience store sells ##x## bottles of sunscreen and ##x^2## pints of ice cream.
We could write the algebra relationship like this:
And it’s correct… but misleading!
The equation implies...
We calculate probabilities based on what we know and not on what more we could know.
Suppose we are asked to choose any number between 1 to 100. One correct number wins. So, for all we know, we have a ##\frac{1}{100}## chance of winning. But if we knew that the person, who designed the game...
Maybe the law of total probability can be used.
A= put a mine in block E.
##E_1## = put a mine in a random block on road AB
##E_2## = put a mine in a random block on road CD
The person who put the mines could have made any decision between ##E_1## and ##E_2##. So...
I think the person who put the mines could have followed this rule:
Number the blocks on road A from 1 to 6, up to down.
Number the blocks on road B from 1 to 13, left to right.
Block E is numbered 2 on road A and 7 on road B.
Now, roll two die simultaneously. Die P has 6 faces. Die Q has 13...
In your example, the probability that you picked 3 is 0.
I've inserted a picture and have written two rules about how the mines were placed:
1. There's only one mine in road AB in one of its blocks.
2. There's only one mine in road CD in one of its blocks.
And we know the number of blocks in...
I thought of this problem:
The roads AB and CD have block E in common. There are 6 blocks in road AB and 13 blocks in road CD.
Someone has planted a mine in some of the blocks. He gives us this information:
1. There's only one mine in road AB in one of its blocks.
2. There's only one mine...
I'm just saying that ##\frac{dp}{ds}## can also be a measure of how hard you push or pull. If you push something hard, it's momentum will change by a larger amount in a small displacement. If you push something with feeble effort, then, for the same change in momentum, the body will have to move...
What? My integral was
$$I=\int_0^v\frac{m}{v\sqrt{1-\frac{v^2}{c^2}}}dv$$
In the link,
$$I=\int_0^v\frac{m}{v(1-\frac{v^2}{c^2})^{\frac{3}{2}}}dv$$ is evaluated
I don't know what ##\left( {\nabla _s \cdot p^T } \right)^T## means. So, I think my calculation is limited to one dimension. I've a small book which used this approach to derive mass-energy equivalence. I'm just doing the same:
$$KE=\int_0^sFds=\int_0^vvd(\gamma mv)$$...
So, can we take the limit of the second integral as lower limit tends to zero and say that:
$$\int_0^t\frac{dp}{ds}dt=\frac{m}{\sqrt{1-v^2}}-m+m\log{\frac{v}{2}}$$
And,why are you putting ##c=1##?
I thought that since ##m\ln v## looks something analogous to energy, so I thought setting ##v=0## in the relativistically correct expression of ##m\ln v## would give me something analoguous to the mass-energy equivalence. Because, suppose name the quantitity ##\frac{dp}{ds}=force'##. And, we...
I'm just trying to get a relativistically correct expression of ##\int_0^t\frac{dp}{ds}dt##
The non-relativistic formula is:
$$\int_{t_1}^{t_2}\frac{dp}{ds}dt=\int_{v_1}^{v_2}m\frac{dv}{v}$$
$$=m(\ln{v_2}-\ln{v_1})$$
Note that this is similar to the work-energy equation...
I tried to get a relativistically correct expression of ##\int_0^t\frac{dp}{ds}dt## similar to the derivation of relativistic energy expression but I got a result which is not defined:
$$\int_0^t\frac{dp}{ds}dt=\int_0^v\frac{d\left(\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}\right)}{v}$$...
If you don't like the name, then maybe use ##force'##. But why doesn't it matter? I've shown that we can also do calculations with it and predict the future of a system.
##\frac{dp}{dt}## is given the name 'force' but ##\frac{dp}{ds}## has no name. I know 'force' is useful for calculations and predicting the future of the system. If 'convenience in calculations' is the reason why some quantities are given names, then I don't see why ##\frac{dp}{ds}## doesn't...
I tried to read it again, but it's too hard to read something when you don't understand more than half of the terms. He said that it involves college level maths. So, I've it saved and I'll read it again when I get to that level. I know that, in the paper, he proves that functional-roots exist...
Could there be a family of functions, one for each point, such that ##f(f(x))=log_ax##?
For example, by using a first degree polynomial, I got this function which when applied twice to ##c## gives you ##a^c##:
$$x\sqrt{a^c\log_ea}+\frac{a^c(1-c\log_ea)}{\sqrt{a^c\log_ea}+1}$$
Problem is it...
I think we can get exact functions ##f(x)## such that ##f(f(x))=logx## using limits. If we assume ##f(x)## to be of degree ##n##. Then ##f(f(x))## will be of degree ##n^2##. Now, we ignore the terms containing powers of ##x## greater than ##n##, and then we solve ##n+1## equations in ##n+1##...
Yeah, I know that. That's why I replaced ##h## in terms of ##n## to get an expression in a single variable before differentiating. However, if the function we're talking about is everywhere positive from a to b then certainly the starting and end terms of
$$\lim_{h\rightarrow...
Obviously the definite integral of a function from a to b is its average value between a and b multiplied by the length of the interval. Why are you talking about that? I thought you were saying that about my final expression.
I think my sum would be undefined in that case. Because remember that it'll be a sum of an infinite number of 0's. I think that's an indeterminate form. Forget about adding the derivatives of ##f(x)##. Instead in my last expression, it says that the double integral of f(x) from a to b is given...
Oh, I guess I had divided by ln(2) instead of 2ln(2) earlier. I always do elementary mistakes even with a calculator. I think you're right. These equations don't work in most cases.
I remember reading in my book that all the other indeterminate forms can be reduced to ##\frac{0}{0}##. So, ##\frac{\infty}{\infty}## is the same as ##\frac{0}{0}##. And, I'm not saying that my last expression is a surprise. The series is not making any sense to me. How is it the average value...
The definite integral of a function ##f(x)## from ##a## to ##b## as the limit of a sum is:
$$\int_a^bf(x)dx=\lim_{h\rightarrow 0}h(f(a)+f(a+h)+.. ..+f(a+(n-2)h)+f(a+(n-1)h))$$
where ##h=\frac{b-a}{n}##. So, replacing ##h## with ##\frac{b-a}{n}## gives:
$$\lim_{n\rightarrow...
And the formula should certainly work if there's no bug in this:
$$\text{Si}\left(x\right)=\text{Si}(a)+\frac{(x-a) \sin (a)}{a}+O\left((x-a)^2\right)$$
Then, composing series
$$\text{Si}\left(B^2\right)=\text{Si}\left(A^2\right)+\frac{2 (B-A) \sin...
I already told you that ##logA## and ##logB## should be close, which would make the fractional large even if the numerator is small. Since the graph of logarithm almost becomes parallel to the x axis, so, the formula would work even when A and B differ by large amounts. And, I checked for A=1...
Let's talk about your definition of force, which appears to be ##mass*distance moved*roughness of surface## . Apply the same 'effort' on two blocks of same mass on two surfaces of same roughness, one on Earth and one on the Moon, the same block will move through different distances. So, this...
In a similar way, I got the general rule:
A function ##f(x)## is differentiable ##n## times only if these two limits are equal:
1.$$\lim_{h\rightarrow 0}\frac{\sum_{r=0}^n(-1)^r\cdot ^nC_r\cdot f(x+(n-r)h)}{h^n}$$
2.$$\lim_{h\rightarrow 0}\frac{\sum_{r=0}^n(-1)^r\cdot ^nC_r\cdot...
Sorry, I mistakenly reported my own post last time. But later I realized that these limits do work. So, I'm posting this again.
I'm using these limits to check second-order differentiability:
$$\lim_{h\rightarrow 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$
And,
$$\lim_{h\rightarrow...
I've been thinking about it since yesterday and have noticed this pattern:
We have, the first order derivative of a function ##f(x)## is:
$$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} ...(1)$$
The second order derivative of the same function is:
$$f''(x)=\lim_{h\rightarrow...
I fully appreciate your effort. I don't know advanced programming concepts yet. I haven't even started arrays. So, I thought instead of working on my skills for months and then checking my work, I'd ask Stephen Tashi to quickly check it only if it wasn't too much work. As he later mentioned that...
I'm not sure but I believe the actual requirement of the formula is ##logA## should be close to ##logB## instead. Could you check it for ##A=100## and ##B=110## by a computer program. I'd have to evaluate too many terms for that.