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!

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

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.

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 :)

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!

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.

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
–
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 \[ \frac{4 {48 \choose 12}}{52 \choose 13} \]

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
–
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.

@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.

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?

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
–
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
–
This is what i'm talking about !{http://en.wikipedia.org/wiki/Lawofcosines#Vector_formulation} . And thanks for clearing the second doubt . :)

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).

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??

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

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.

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)

@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)

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,

Yes I know this but what is the meaning of this expression we get after diffrentiating-\(\dfrac{tangent ~of ~f(a)}{tangent ~of ~f(b)} ~ or ~ \dfrac{rate ~ of~change ~of ~f(a)}{rate~of~change ~of ~f(b)}\)

\(\lim_{ x \to \infty} f(x)^{g(x)}\) 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.

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 \(\dfrac{tangent ~of ~f(a)}{tangent ~of ~f(b)} ~ or ~ \dfrac{rate ~ of~change ~of ~f(a)}{rate~of~change ~of ~f(b)}\)

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

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

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

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.

## Comments

Sort by:

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

s

Log in to reply

Wow! Nice picture.

Log in to reply

Code credit: @Agnishom Chattopadhyay

Log in to reply

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

Log in to reply

It has no complex solution, either.

Log in to reply

In what ways do you consider yourself strange?

Log in to reply

... in strange ways, of course.

Perhaps, I am strange in that I

want to be strange.Log in to reply

me too

Log in to reply

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

NOto all those sequence problems by proving that there are multiple answers? Please share! RIGHT HERE!And Happy New Year!

Log in to reply

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.

Log in to reply

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

Log in to reply

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 :)

Log in to reply

Interesting Question.

Agni = Fire Shom = Moon

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

:)

Log in to reply

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

\(\ddot\smile\)

Log in to reply

Log in to reply

Hey Agnishom how's you doin?

Have you eliminated the need for intellectual thought yet? ;)

Log in to reply

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

I'm doing fine.

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

Log in to reply

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

Log in to reply

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:

StatementGiven 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 = CProofSuppose 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.

Log in to reply

Thanks nicely explained , so it just provides us flexibility.

Log in to reply

Log in to reply

s

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 :)

Log in to reply

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!

Log in to reply

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.

Log in to reply

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

Thanks for sharing.

Log in to reply

When did you start doing math. At what age did you really start to learn math outside of school.

Log in to reply

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

Log in to reply

a

s

sss

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

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.

Log in to reply

Please solve all my doubts on this

Log in to reply

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 \(\ddot\smile\)

Log in to reply

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

The second answer is probably 9139/20825

Log in to reply

Could you elaborate ?

Log in to reply

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 \[ \frac{4 {48 \choose 12}}{52 \choose 13} \]

Log in to reply

Log in to reply

Log in to reply

Check tis out!

Please participate and reshare.

Log in to reply

That is nice.

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

Log in to reply

Do reshare it. Let everyone participate. I liked it very much and reshared it. You should do the same.

Log in to reply

Log in to reply

Log in to reply

Comment deleted Mar 08, 2015

Log in to reply

Comment deleted Mar 08, 2015

Log in to reply

Log in to reply

Log in to reply

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

Log in to reply

@Rajdeep Dhingra did. He asked everyone to tell it to others so more people can participate. That's what I was doing.

I haven't organized this . ThatLog in to reply

Anyway, question 3 needs to be rephrased for clarity

Log in to reply

@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.

image

Log in to reply

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?

Log in to reply

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 ?

Log in to reply

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

Log in to reply

ofcosines#Vector_formulation} . And thanks for clearing the second doubt . :)Log in to reply

please solve! https://brilliant.org/problems/how-to-make-a-triangle/?group=ZTtB5mdNQtMx

Log in to reply

fixed.

Log in to reply

please write a solution

Log in to reply

check this

I do not know how to solve it, yet butLog in to reply

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??

Log in to reply

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?

Log in to reply

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).

Log in to reply

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??

Log in to reply

You wanted a proof other than this?

Log in to reply

Log in to reply

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

Log in to reply

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

canactually still have a box with 0 things.Log in to reply

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

Log in to reply

@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,

Log in to reply

Yes I know this but what is the meaning of this expression we get after diffrentiating-\(\dfrac{tangent ~of ~f(a)}{tangent ~of ~f(b)} ~ or ~ \dfrac{rate ~ of~change ~of ~f(a)}{rate~of~change ~of ~f(b)}\)

Log in to reply

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

can you explain what you mean by

here we directly integrate the function?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

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

\(\lim_{ x \to \infty} f(x)^{g(x)}\) 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.Log in to reply

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 \(\dfrac{tangent ~of ~f(a)}{tangent ~of ~f(b)} ~ or ~ \dfrac{rate ~ of~change ~of ~f(a)}{rate~of~change ~of ~f(b)}\)

Log in to reply

I will write a note on it. What past?

estands foreternalI'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 partsis the same asfinding the anti-derivativeis called theFunamental Theorem of Calculus.Explanation 2: I could fill up the area under the curve like this.

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

integrateorsum up all the small stripsto 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.

Log in to reply

That are small strips, just for clear visibility made large.

Log in to reply

Anyhow, could I answer your query about integrals successfully?

Log in to reply

solve this https://brilliant.org/problems/can-you-solve-it-9/?group=j4P7dyAoNhAb&ref_id=568173

Log in to reply

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

Log in to reply

another one https://brilliant.org/problems/train-your-brain/?group=Dm8rRm81UDqj

Log in to reply