Agnishom's Messageboard

Wow! Nice picture.

I was going to answer negative infinity but no. It has no real solution.

... in strange ways, of course.

Perhaps, I am strange in that I want to be strange.

Interesting Question.

Agni = Fire Shom = Moon

Agnishom = Two (Condradictory) Characters at once, especially highlighting the active and the passive

:)

Oh sorry, I went to bed last night because my head was aching terribly.

Happy GNU Year!

One way could be looking the OEIS. Or you could fit a number of curves through the points.

Sorry @John Muradeli can you post all your notes, problems in simple english , sometimes I can't understand as I am poor in english and too a gujarati

Age 3. My parents taught me to count at age 3 which they taught in school when I was 4.

I think the quote is by Alfred N. Whitehead (not sure who he is, though)

I got the quote from A=B

Well, in Mathematics we are constantly trying to generalize out things. Even though Mathematicians are very intelligent, a discovery is valued higher if it can be utilised without intelligent thought.

For example, I'd say that the method of solving an equation by using the quadratic formula is a 'better' discovery than the method of middle term factorisation because that can make anyone solve a quadratic equation - even without understanding what it means!

A fool, no questioner is. By asking, wiser he becomes. -Yoda

It is not a poor doubt. It is a good question.

Projections are dependent upon the direction you look from. You usually project an inclined vector on the x-axis by dropping its perpendicular because you're interested in 'splitting it into orthogonal components'. 

When your axes are inclined at an angle $\theta$, you'd be not be dropping perpendiculars. You'd be dropping components that touch the other axes at $\theta$.

Ok, your main question is whether it'd be possible to have axes not perpendicular to each other without breaking things.

The answer is yes, you'd still be able to represent each point uniquely. 

Look at the following theorem:

Statement Given any two non-collinear vectors A and B, and another vector C, all on the same plane, there exists unique scalars a and b such that aA + bB = C

Proof Suppose not. Then, we'd have scalars x and y that'd give me xA+ yB = C.

By hypothesis, we have: aA + bB = xA + yB Or, aA - xA = yB - bB Or, (a-x)A = (y-b)B Or, ((a-x)/(y-b))A = B

So, it looks like there is a scalar which when multiplied with the first vector, gives the second vector. That is only possible if the two vectors shared the same direction.

But wait, we defined our vectors to be in different directions. Contradiction!

So, a and b must be unique.

You could think of the coordinates of the point as a position vector, just like C and compare A and B with the axes.

Now, you should be able to see that how the theorem still holds.

Okay, a picture to make it look believable.

Imgur

Is there a good reason to choose $\frac{\pi}{2}$ as the angle between the axes?

There are many. The most important one of them is that it allows us to use the pythagorean theorem to find the distance between two points immediately.

It cannot be done but we are approaching that point. (asymptotes?)

I'm doing fine.

Christmas Wishes! How're your AP Classes going on?

I need some time to answer the third question. But it is a good question and worth knowing.

1. can you explain what you mean by here we directly integrate the function?

2. Suppose that f(c) = g(c) = 0

$\lim_{x\to c}\frac{f(x)}{g(x)} \\ = \lim_{x\to c}\frac{f(x)-0}{g(x)-0} \\ = \lim_{x\to c}\frac{f(x)-f(c)}{g(x)-g(c)} \\ = \lim_{x\to c}\frac{\left(\frac{f(x)-f(c)}{x-c}\right)}{\left(\frac{g(x)-g(c)}{x-c}\right)} \\ =\frac{\lim\limits_{x\to c}\left(\frac{f(x)-f(c)}{x-c}\right)}{\lim\limits_{x\to c}\left(\frac{g(x)-g(c)}{x-c}\right)}$

Now, suppose that h = x - c

$= \frac{\lim\limits_{x\to c}\left(\frac{f(c+h)-f(c)}{h}\right)}{\lim\limits_{x\to c}\left(\frac{g(c+h)-g(c)}{h}\right)} \\ = \frac{f'(c)}{g'(c)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}$

1. Clearing some confusions:

• $\lim_{ x \to \infty} f(x)^{g(x)}$ where $\lim_{ x \to \infty} f(x) = 1$ $\lim_{ x \to \infty} g(x) = \infty$ and is an indeterminate form. What this means is that, the limit cannot be calculated straight away from this expression. It conceals a value within. It is not necessarily undefined. For Example, $\lim_{n\to\infty}\left(1+\frac1n\right)^n=e\approx2.718281828459045$

• $1^{\infty}$ is not defined. This is because infinity is not really a real number. It is a limiting idea.

Now, for that limit, it evaluates to 1 (for obvious reasons). This can be rigorously proven or shown with a graph.

You should think of limits as a graph/function approaching a value, not an abstract algebraic operation.

@megh choksi - I have attempted to answer your "why we differentiate" question and heres what i got, just check it out.

Note that it is not very rigorious or perhaps even entirely correct, and i suggest you to check out the proof of L'hospital which comes from Rolles theorem,

Suppose you have two functions f(x) and g(x),, and for simplicity suppose that they converge at (0,0)

Then,

$\\ Lim_{ h\quad tends\quad to\quad 0 }\frac { f(h) }{ g(h) } =a\\$

so,

$f(h)=a(g(h))$

For really small 'h'

Now consider the tangent at origin,

Then the tangent at any point is given by ,

$f'(x)=\frac { y-{ y }_{ 0 } }{ x-{ x }_{ 0 } } =\frac { y }{ x }$

and similarly

$g'(x)=\frac { y }{ x } \quad$

Now for a point close, to 0, say 'h'

we have approximately,

$f'(x)=\frac { f(h) }{ h } \quad$

$g'(x)=\frac { g(h) }{ h } \quad$

where h is arbitarily close to 0,

Now divide both of them ,

$\frac { f'(x) }{ g'(x) } =\frac { f(h) }{ g(h) } \\$

We see that the ratio of tangents is almost same as ratio of the functions themselves at limiting point, and so ratio of derivatives also give us the same limit

a

Hence, proved

Note it is not valid in cases where we cant define a tangent at origin, which is almost all the cases,, this is just what i got, and i dont know how valid it is, any mathematician would only get angry at what i did i suppose, but it does provide a qualitative idea, about how the ratio of slopes becomes equal to the ratio of functions limiting values,

11 year kids are not idiots.

I'll show him some analogy.

What happens when I choose one object out of four objects A,B,C,D and keep them in a box? The new box looks like any one of this {A},{B},{C},{D}. How many boxes can I have? 4

What if I choose two objects? I can then have {A,B},{A,C},{A,D},{B,C},{B,D},{C,D}. How many boxes can I have? 6

What if I choose no objects? What does the box look like? It is just an empty box - {}. So, I can actually still have a box with 0 things.

I cannot exactly explain this but I can prove this. So, you'll get a semi-satisfactory answer.

It appears that you are not aware of Modular Arithmetic which you should know to understand the following proof:

Consider the trivial case: $6 \times 2 \equiv 2 \pmod {10}$

Now, supposing that the following is true:

$6 \times 2n \equiv 2n \pmod {10}$

We have:

$(6 \times 2n)+n \equiv 2n+2 \pmod {10} \\ \implies 6 \times 2n)+12 \equiv 2n+12 \pmod {10} \\ \implies 6 \times 2(n+1) \equiv 2n + 2 \pmod {10} \\ \implies 6 \times 2(n+1) \equiv 2(n+1) \pmod {10}$

Thus by induction we have $6 \times 2k \equiv 2k \pmod {10}$ $\qed$

EDIT: You're level 5 in Number Theory. You probably know modular arithmetic. Why did you ask the question, then?

fixed.

Hi;

I had not noticed this earlier. So, sorry for the late reply.

1) The cosine rule helps you draw the third side of a triangle. However, when you have two co-initial vectors making an angle with each other, you want the diagonal of the parallelogram containing those two vectors, not the third side. However, the formulae are closely related. Guess why?

That is nice.

Will participate if I feel like it. I'm feeling more computerish today

Is the answer to the first question: 6327/20825 ?

The second answer is probably 9139/20825

This one is not so tough. Try integrating the first term wrt x, the second wrt y