Maths is Fun

Screambowl

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Hi, I am starting a new thread which deals with discussing maths both in India and abroad.

This thread is not just related to discussion of history of maths but also discussing actual mathematical problems. You may post your queries and if any one can solve then most welcome :D
 

Screambowl

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''Sridhara was an Indian mathematician who wrote on practical applications of algebra and was one of the first to give a formula for solving quadratic equations''

there is a long derivation which comes to a conclusion with discriminant formula D=b²-4ac and there fore finding both solutions.

In 628 AD, Brahmagupta, an Indian mathematician, gave the first explicit (although still not completely general) solution of the quadratic equation ax2 + bx = c as follows: "To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the value." (Brahmasphutasiddhanta, Colebrook translation, 1817, page 346)[16]:87 This is equivalent to:
 

Screambowl

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upload_2017-8-3_9-26-33.png


here m is the slope.

slope is nothing but a tangent to a curve (, parabolic, hyperbolic, exponential function, etc). Which describes how steep the graph is going.

1)In case you want to find a slope of a line then it's y2-y1/x2-x1 where (x1,y1) (x2,y2) are two points on the line.

2) in case you want to find the slope of a parabolic curve or other polynom (high order of a degree of fucnt.) I.e. path of a projectile or missile, then find the derivative of it. The first derivative will be the slope.
 

Adioz

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Hi, I am starting a new thread which deals with discussing maths both in India and abroad.

This thread is not just related to discussion of history of maths but also discussing actual mathematical problems. You may post your queries and if any one can solve then most welcome :D
Nice thread. I am extremely weak in Maths. Someone help me solve this:-
Complex Integral query.png


Answer = 0
 

Adioz

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aditya10r

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Here for help.

COS^2x=(1+cos^2x)/2 and Sin^2x=(1-cos^2x)/2
 

aditya10r

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Switzerland :D ana to wapis hi hai but yahan maths bohot phaduu hai..
...................................................
What do they teach their dude???

How to find the inverse of a 4x4 matrix??
 

Adioz

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:facepalm:Kitna chutiya hoon main?

Cos 2z is analytic.


Direct integration works. I was trying to substitute z=π*(e^iθ) with θ limits from -π/2 to π/2. From there it went way too complex (pun intended).

OTOH direct integration gave me answer = 4 Sin2πi

BTW, I was also seeing the wrong answer :doh:. The answer was not zero. I saw the answer of exercise 14.2 instead of 14.1
 

Screambowl

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What do they teach their dude???

How to find the inverse of a 4x4 matrix??
that's basic ... taught in 12th Bhai :p ...
.........................................................
 

Adioz

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that's basic ... taught in 12th Bhai :p ...
.........................................................
They teach Gauss-Jordan elimination in 12th?o_O
 

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