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Note by Agnishom Chattopadhyay
2 years, 7 months ago

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\[\left| x \right| < 0\] \[x \in ?\]

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John Muradeli · 2 years, 7 months ago

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@John Muradeli Wow! Nice picture. Agnishom Chattopadhyay · 2 years, 7 months ago

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@Agnishom Chattopadhyay

Code:

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def f(z):
    x = real(z)
    y = imaginary(z)
    return (x^2 + y^2 -1)^3 - x^2*y^3

complex_plot(lambda z: (1/f(z)*i^(z+1)),(-1.5,1.5),(-1.5,1.5),plot_points=300)

Code credit: @Agnishom Chattopadhyay John Muradeli · 2 years, 7 months ago

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@John Muradeli I was going to answer negative infinity but no. It has no real solution. Les Narvasa · 2 years, 7 months ago

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@Les Narvasa It has no complex solution, either. Agnishom Chattopadhyay · 2 years, 7 months ago

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In what ways do you consider yourself strange? Calvin Lin Staff · 2 years, 7 months ago

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@Calvin Lin ... in strange ways, of course.

Perhaps, I am strange in that I want to be strange. Agnishom Chattopadhyay · 2 years, 7 months ago

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@Agnishom Chattopadhyay me too Megh Choksi · 2 years, 4 months ago

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My computer is being a pain right now so I can't quite "reply" but I'll reply here:

AP Classes: NOPE! DON'T MENTION!

Alright well I wanted to tell you you officially become my hero when you made that post disputing all the sequence problems (that one that went 1,2,3,4,5,6,7,8,9,__ and the comic started talking some geometric gibberish and primes from \(n^2-11n+1\) or something).

So, is there a general way to simply say NO to all those sequence problems by proving that there are multiple answers? Please share! RIGHT HERE!

And Happy New Year!

John Muradeli · 2 years, 4 months ago

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@John Muradeli 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. Agnishom Chattopadhyay · 2 years, 4 months ago

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@John Muradeli 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 Megh Choksi · 2 years, 4 months ago

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I'am just curious to know that what is meaning of your name ? because i always feel that it is difficult to read , since it is at least new for me :) Deepanshu Gupta · 2 years, 5 months ago

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@Deepanshu Gupta Interesting Question.

Agni = Fire Shom = Moon

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

:) Agnishom Chattopadhyay · 2 years, 4 months ago

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@Agnishom Chattopadhyay Wow That Interesting ! It Means You Like All Seasons as you have Power to remain Cool in summer and Hot in winter

\(\ddot\smile\) Deepanshu Gupta · 2 years, 4 months ago

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@Deepanshu Gupta Somebody explained it to me (jokingly) as "The Moon burnt dark in Fire" Agnishom Chattopadhyay · 2 years, 4 months ago

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Hey Agnishom how's you doin?

Have you eliminated the need for intellectual thought yet? ;) John Muradeli · 2 years, 4 months ago

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

I'm doing fine.

Christmas Wishes! How're your AP Classes going on? Agnishom Chattopadhyay · 2 years, 4 months ago

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Have a poor doubt - Why do we take the axes always perpendicular . Is it so because we are treating it as 2 independent scales, if they are at an angle then there would be a projection of one on other , making it perpendicular , it has its projection on its own thus making it independent @Agnishom Chattopadhyay Megh Choksi · 2 years, 5 months ago

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@Megh Choksi

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

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. Agnishom Chattopadhyay · 2 years, 5 months ago

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@Agnishom Chattopadhyay Thanks nicely explained , so it just provides us flexibility. Megh Choksi · 2 years, 5 months ago

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@Megh Choksi Yep Agnishom Chattopadhyay · 2 years, 5 months ago

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Is this quote your original?

I'd like to know what it means, or what you think it means.

I like big-idea quotes, but a lot of times I observe that people who came up with them have completely different explanations and justifications to the relative respects of applications for which the quote applies; so I thought I'd hear you out :) John Muradeli · 2 years, 7 months ago

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@John Muradeli 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! Agnishom Chattopadhyay · 2 years, 7 months ago

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@Agnishom Chattopadhyay Whitehead was a great philosopher and mathematician, I believe; he worked alongside the great Bertrand Russell in writing the Principia Mathematica. And then he went insane. Jake Lai · 2 years, 4 months ago

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@Agnishom Chattopadhyay Well, in this case, I'm an anti-thesis as far as my position goes for this quote.

Thanks for sharing. John Muradeli · 2 years, 7 months ago

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When did you start doing math. At what age did you really start to learn math outside of school. Trevor Arashiro · 2 years, 7 months ago

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@Trevor Arashiro Age 3. My parents taught me to count at age 3 which they taught in school when I was 4. Agnishom Chattopadhyay · 2 years, 7 months ago

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@Agnishom Chattopadhyay

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John Muradeli · 2 years, 7 months ago

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@John Muradeli That's the case with most Indians. They learn counting around the age of 3 . :) Keshav Tiwari · 2 years, 5 months ago

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@Keshav Tiwari And learning about life at around the age of 40. John Muradeli · 2 years, 5 months ago

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@John Muradeli Don't know about that , Still 17 ! :) :D Keshav Tiwari · 2 years, 5 months ago

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@John Muradeli that's really funny! Les Narvasa · 2 years, 7 months ago

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@John Muradeli We all learn to count at 3. Agnishom Chattopadhyay · 2 years, 7 months ago

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@Agnishom Chattopadhyay I learnt counting when I was \(\pi\) years old. Satvik Golechha · 2 years, 7 months ago

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@Satvik Golechha I didn't know what numbers were when I was \(2\phi\) Trevor Arashiro · 2 years, 7 months ago

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@Trevor Arashiro I still at the age of \(e \times \phi \times \pi\) don't know what numbers are, though I've started exploring what they do. Satvik Golechha · 2 years, 7 months ago

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@Satvik Golechha ahhh yes, the deep philosophical question of what are numbers, I think I'll write a paper on that one day. Trevor Arashiro · 2 years, 7 months ago

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@Satvik Golechha Well, I don't know what numbers are, either. Agnishom Chattopadhyay · 2 years, 7 months ago

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@Agnishom Chattopadhyay The Alphabet is to Literature what numbers are to maths. Satvik Golechha · 2 years, 7 months ago

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@Satvik Golechha Wow! Isn't it amazing that these constant product to ~\(14\). :D! Krishna Ar · 2 years, 7 months ago

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@Krishna Ar You'll be more surprised to calculate the value of \(e^{\pi} -{\pi}\) Agnishom Chattopadhyay · 2 years, 7 months ago

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@Agnishom Chattopadhyay Woah! That was cool, for once I thought it was exactly 20, though later I realized that it can't be. It's 19.9990999792. Satvik Golechha · 2 years, 7 months ago

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@Satvik Golechha BTW How did you find it out? Bash?? Satvik Golechha · 2 years, 7 months ago

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@Satvik Golechha No, it's from xkcd. Agnishom Chattopadhyay · 2 years, 7 months ago

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@Krishna Ar Isn't it amazing enough that these constants are amazing? Satvik Golechha · 2 years, 7 months ago

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@Agnishom Chattopadhyay This is supposed to a compliment. As in "DARN THOSE SMART KIDS!"

It's internet stuff... Only those who do it get it. But yeah, the "we" you're talking about encompasses only a small portion of the global population.

Good for you. John Muradeli · 2 years, 7 months ago

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Please solve all my doubts on this Rajdeep Dhingra · 2 years, 1 month ago

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I can't believe I wasn't able to solve this.
A person picked up 13 cards from 52 cards. Find Probability that he has no ace , Probability that he has exactly one ace ?
Please solve this \(\ddot\smile\) Rajdeep Dhingra · 2 years, 2 months ago

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@Rajdeep Dhingra Is the answer to the first question: 6327/20825 ?

The second answer is probably 9139/20825 Agnishom Chattopadhyay · 2 years, 2 months ago

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@Agnishom Chattopadhyay Could you elaborate ? Rajdeep Dhingra · 2 years, 2 months ago

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@Rajdeep Dhingra There are \( 52 \choose 13 \) combinations of 13 cards.

There are \( 48 \) cards which are not aces. Hence, there are \(48 \choose 13\) combinations of non-aces.

So, the probability that you will run into one such configuration is \[ \frac{48 \choose 13}{ 52 \choose 13 } \]

Given any set of \(48\) cards, none of which are aces, there are \( 48 \choose 12 \) combinations of 12 cards.

To create a set of \(13\) cards, exactly one of which is an ace, one has \(4\) options to choose the ace and \( 48 \choose 12 \) options to choose the rest of the cards.

Hence, the probability you are looking for is \[ \frac{4 {48 \choose 12}}{52 \choose 13} \] Agnishom Chattopadhyay · 2 years, 2 months ago

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@Agnishom Chattopadhyay Thanks ! Rajdeep Dhingra · 2 years, 2 months ago

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@Rajdeep Dhingra you're welcome! Agnishom Chattopadhyay · 2 years, 2 months ago

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Check tis out!

Please participate and reshare. Rajdeep Dhingra · 2 years, 2 months ago

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@Rajdeep Dhingra That is nice.

Will participate if I feel like it. I'm feeling more computerish today Agnishom Chattopadhyay · 2 years, 2 months ago

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@Agnishom Chattopadhyay Do reshare it. Let everyone participate. I liked it very much and reshared it. You should do the same. Tanveen Dhingra · 2 years, 2 months ago

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@Tanveen Dhingra I've reshared it. Agnishom Chattopadhyay · 2 years, 2 months ago

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@Agnishom Chattopadhyay Okay.....that's good Tanveen Dhingra · 2 years, 2 months ago

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Comment deleted Mar 08, 2015

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Comment deleted Mar 08, 2015

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@Tanveen Dhingra Sorry, I apologise. Agnishom Chattopadhyay · 2 years, 2 months ago

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@Agnishom Chattopadhyay It's okay. I too apologise for replying in a rude way. sorry

Tanveen Dhingra · 2 years, 2 months ago

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@Tanveen Dhingra It's alright.

While I appreciate your efforts in organising this, I might not actively participate since whether I am good at number theory at a moment depends on things like the weather, my emotions, etcetera Agnishom Chattopadhyay · 2 years, 2 months ago

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@Agnishom Chattopadhyay I haven't organized this . That @Rajdeep Dhingra did. He asked everyone to tell it to others so more people can participate. That's what I was doing. Tanveen Dhingra · 2 years, 2 months ago

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@Tanveen Dhingra I thouught you're brothers.

Anyway, question 3 needs to be rephrased for clarity Agnishom Chattopadhyay · 2 years, 2 months ago

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@Agnishom Chattopadhyay I have some doubts . It would be great help if you take a look at them : 1)As per the cosine rule \(a^2=b^2+c^2-2bccos(A) \) where symbols have their usual meaning .But , in vector addition the magnitude of the sum of two of two vectors b and c placed in order and making an angle A is \(b^2+c^2+2bccos(A)\) Why is it so ? \(\quad \quad \quad 2)\) I encountered a question in which a cylinder floats at the interface of two liquids of density \(d_{1}\) and \(d_{2}\) with mass of cylinder as A and height h/3 in \(d_{2}\) and 2h/3 in \(d_{1}\) . the equation of the forces can be written as \(mg-d_{1} (2h/3)Ag- d_{2}(h/3)Ag=0 \). Now where does the buoyant force due to liquid \(d_{1}\) act . I mean does as like a suction as there's no space from bottom area.

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Please help ! Keshav Tiwari · 2 years, 4 months ago

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@Keshav Tiwari 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? Agnishom Chattopadhyay · 2 years, 3 months ago

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@Agnishom Chattopadhyay Thanks for replying. I just read that the law of cosines is equivalent to the dot product of two vectors. Is that the correlation you're talking about ? Keshav Tiwari · 2 years, 3 months ago

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@Keshav Tiwari I don't understand why the law of cosines is equivalent to the dot-product.

2) no, there is no suction like force that pulls up the upper part. What actually happens is like this: That part displaces the liquid around it which presses against the liquid below which gives the extra force through the bottom of the cylinder. In other words, all the force acts from down under the cylinder but the nett result is what you wrote Agnishom Chattopadhyay · 2 years, 3 months ago

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@Agnishom Chattopadhyay This is what i'm talking about !{http://en.wikipedia.org/wiki/Lawofcosines#Vector_formulation} . And thanks for clearing the second doubt . :) Keshav Tiwari · 2 years, 3 months ago

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please solve! https://brilliant.org/problems/how-to-make-a-triangle/?group=ZTtB5mdNQtMx Visakh Radhakrishnan · 2 years, 4 months ago

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@Visakh Radhakrishnan fixed. Agnishom Chattopadhyay · 2 years, 4 months ago

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@Agnishom Chattopadhyay please write a solution Visakh Radhakrishnan · 2 years, 4 months ago

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@Visakh Radhakrishnan I do not know how to solve it, yet but check this Agnishom Chattopadhyay · 2 years, 4 months ago

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When you multiply 6 by an even number, they both end in the same digit.

Example: 6×2=12, 6×4=24, 6×6=36, etc

Why does this happen?? Harshvardhan Mehta · 2 years, 4 months ago

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@Harshvardhan Mehta 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? Agnishom Chattopadhyay · 2 years, 4 months ago

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@Agnishom Chattopadhyay I have a different proof.

We have to prove that

\[6 \times 2n \equiv 2n \pmod{10}\]

This is equivalent to

\[(5+1) \times 2n \equiv 2n \pmod{10}\]

\[10n + 2n \equiv 2n \pmod{10}\]

\[ 2n \equiv 2n \pmod{10}\]

This result is true. Using this same method we can prove that units digit of 6 times an odd number is the units digit of (units digit of odd number + 5). Pranshu Gaba · 2 years, 4 months ago

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@Agnishom Chattopadhyay well i just wanted to know if there were any other ways of proving or displaying this.. anyways thanx for ur proof btw u really a 16 year old?? if yes u preparing for JEE?? Harshvardhan Mehta · 2 years, 4 months ago

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@Harshvardhan Mehta I'm preparing for JEE but it's not likely that I'll study engineering.

You wanted a proof other than this? Agnishom Chattopadhyay · 2 years, 4 months ago

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@Agnishom Chattopadhyay no i mean u have superb knowledge buddy keep it up.. Harshvardhan Mehta · 2 years, 4 months ago

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How do you clear your doubts?

Now suppose consider me as a 11 year dumb , idiotic small child-

I open up the first page of P&C chapter , there I see this the number of ways of selecting 0 things from n things is 1 - at very first sight it may seem like a big blunder . So how will you explain this in your way to a 11 year dumb , idiotic small child Megh Choksi · 2 years, 4 months ago

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@Megh Choksi 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. Agnishom Chattopadhyay · 2 years, 4 months ago

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Again some doubts(basics)

1) A problem - \(1\times1\times1\times1\times........=\)

  • \(1\)

  • \(\prod 1\)

    • \(\ not ~ defined\)

    • \(non ~ of ~the~above\)

my answer - \( lim_{x \to \infty} 1^x = ~Thus ~ not~defined\) correct answer was 1! where I was wrong?

2) While evaluating limits get these forms(this only we do in 11th - 12th) - \(\dfrac{0}{0} , \dfrac{\infty}{\infty}\) , we then differentiate the numerator independently and the denominator too, was the meaning of this?

3) What e ? ( I know its euler's constant)

4) You know any logical idea for this - In integration we integrate the integrand(which is the derivative of a well defined function) , now just next step we are studied Application of integrals i.e Area bounded by the curves , here we directly integrate the function, ( I know the geometrical proof of this, but I think you will have a better one)

Thanks , you always helped me Megh Choksi · 2 years, 4 months ago

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@Megh Choksi @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, Mvs Saketh · 2 years, 4 months ago

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@Mvs Saketh Yes I know this but what is the meaning of this expression we get after diffrentiating-\(\dfrac{tangent ~of ~f(a)}{tangent ~of ~f(b)} ~ or ~ \dfrac{rate ~ of~change ~of ~f(a)}{rate~of~change ~of ~f(b)}\) Megh Choksi · 2 years, 4 months ago

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@Megh Choksi 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. Agnishom Chattopadhyay · 2 years, 4 months ago

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@Agnishom Chattopadhyay Sorry sometimes I may be illogical

4) I know this proof(saw in a book)(wonderful) -

\( Ar.(PDFC) < ~Ar.(PQDC)~<Ar.(EQDC)\)

\( f(x)\bigtriangleup x < \bigtriangleup A < f(x + \bigtriangleup x).(\bigtriangleup x)\)

\( lim_{x \to 0} f(x) < lim_{x \to 0} \dfrac{\bigtriangleup A}{\bigtriangleup x} < lim_{x\to 0} f(x + \bigtriangleup x)\)

\( f(x) < \dfrac{dA}{dx}<f(x)\)

as continues

\(\dfrac{dA}{dx} = f(x)\)

\( dA = f(x)dx\)

\(A = \int f(x)dx\)


By directly integrate i mean (take this eg)- \( \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\)

Now for finding the area , we find a expression for y from the equation and \(A =4( \int ydx)\)

I was just asking for a alternative proof(I liked very much your reasoning in the note - An informal yet conceptual note on derivatives) , leave 3 as its related to past

For 2) I was asking what do we mean by this \(\dfrac{tangent ~of ~f(a)}{tangent ~of ~f(b)} ~ or ~ \dfrac{rate ~ of~change ~of ~f(a)}{rate~of~change ~of ~f(b)}\) Megh Choksi · 2 years, 4 months ago

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@Megh Choksi

  1. I do not know of an absolute intuitive explanation for this. However, if you want a geometric proof of the argument, see this.

  2. I will write a note on it. What past? e stands for eternal

  3. I'll try to explain two things but I am not sure what you are exactly asking for.

Explanation 1: Suppose that \(f\) is the derivative of \(F\). Now, this would mean that \(f\) gives the change in the height of the \(F\) curve for a small change in its x-coordinate (which we call \(dx\)). Now, since \(f\; dx\) gives back the height for that part (because slope*change-in-x = change in y, it must equal \(dF\).

By integration what we mean summing up the small parts, which is why we use the \(\int\) sign which is actually a distorted S.

So, it must be that \(\int dF = F\) since F constitutes of the small parts of F. but we just saw that \(dF = f\; dx\). So, we have \(F = {\int f\; dx}\)

This is why you find the anti-derivative while integrating indefinitely. The theorem which shows that summing up the small parts is the same as finding the anti-derivative is called the Funamental Theorem of Calculus.

Explanation 2: I could fill up the area under the curve like this.

integration

integration

The picture should make it clear why a thinner strip works better.

Now, note that each infinitesimally small strip is a rectangle of width \(dx\) and height equal to the value of the function at that point which is \(f(x)\) and thus an area of \(f(x)\;dx\).

As explained earlier, we want to integrate or sum up all the small strips to get the area. So, the area under the curve in the interval [a,b] is \[|\int_{a}^{b}{f(x)\,dx}|\]

@megh choksi , please tell me if this is what you were looking for. Agnishom Chattopadhyay · 2 years, 4 months ago

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@Agnishom Chattopadhyay Thanks ,by past I saying its related to history

That are small strips, just for clear visibility made large. Megh Choksi · 2 years, 4 months ago

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@Megh Choksi Not just small, they are infinitesimal!

Anyhow, could I answer your query about integrals successfully? Agnishom Chattopadhyay · 2 years, 4 months ago

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solve this https://brilliant.org/problems/can-you-solve-it-9/?group=j4P7dyAoNhAb&ref_id=568173 Visakh Radhakrishnan · 2 years, 4 months ago

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@Visakh Radhakrishnan This one is not so tough. Try integrating the first term wrt x, the second wrt y Agnishom Chattopadhyay · 2 years, 4 months ago

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another one https://brilliant.org/problems/train-your-brain/?group=Dm8rRm81UDqj Visakh Radhakrishnan · 2 years, 4 months ago

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