It's knowing when to do that which separates a good mathematician from a plodder... until the plodder has seen so many various series demonstrated that they see the trick and become a good mathematician.
Another way to evaluate A and B in the PFD is by equating coefficients. We have A(x + 1) + Bx ≡ 1 ⇒ (A + B)x + A ≡ 0x + 1 ∴ A + B = 0 equating coefficients in x and A = 1 equating the constant terms ∴ B = -1.
I studieded partial fraction decomposition during my first year at the University (degree in Mathematics), many years ago. I taugh this topic for about 40 years. It is a pleasure mixed with nostalgia to see it reproposed flawlessy. Love and prayers from Italy!
One can also look at the numbers 28, 36, 45, 55, 66, 78,... and notice a pattern that the difference between 2 numbers are in AP with common difference 1. So we can find a formula that generates these numbers by telescopic sum of the difference of the terms Tn - Tn-1, then find a formula for Tn= (1/2)*(n+6)*(n+7). Now if we apply this here, we will get another sum i = 1 to 93 of (2/((i+6)*(i+7))), that gives 2*(1/7 -1/100) = 9/350.
7:49 Can someone please explain why you can set it equal to zero? Wouldn't this mean that you divide 1 by zero on the top-left, which is an undefined move?
Please look at the following link about cover up method ruclips.net/video/fgPviiv_oZs/видео.html 1/(x(x+1)=A/X+B/(X+1) USE COVER UP METHOD A=1/(0+1)=1 B=1/(-1)=-1
Thanks for the video, the 2/2 trick was very creative. I'm a bit confused - in partial fraction decomposition, if x(x+1) = 0, then the fraction 1/((x)(x+1)) is undefined? - dividing by zero. Maybe I am not seeing this correctly.
Consider it this way instead: when you have the equation 1= A(x+1) +Bx, this will be true for all values of x, so you can substitute any value any for x that you want to solve for A and B. The easiest thing to do here is set x to some value so that either A or B is removed from the equation. That would be x=0 (to remove B) or x=-1 (to remove A).
You are very welcome Johnny. You are absolutely correct, denominator must not be zero! However, we used this idea in the PFD! Thank you for your feedback! Cheers! You are awesome 😀
At the point where you set x(x+1)=0 to find A and B, you've already multiplied both sides by x(x+1) and made the denominators go away. So there isn't a problem.
I'll use notebook and pen for your videos.We used to solve HOT questions when we were in 10th. you reminded me the memories of that. I'll use notebook for your videos afterwards
Because greater the k, smaller the 1/k^k is. The sum would look like this 0 + 1/1 + 1/4 + 1/9... And more of them you add up, closer the sum is to 1.292929... but never passes that point.
Unable to believe that i am able to do these questions xD ... wish knew these stuffs last year .. now i am 12th passout and dropper :( (class 13student)
I have a problem and Im wondering if anyone can help me prove it So 12 x 12 is 144 16 x 9 is 144 but 16 x 9 = 13 x 12 And 13 x 12 is 156 Very easy but How?
See that (16-3)(9+3) = (13)(12) but (16- 3)(9+3) is NOT = (16)(9) It is (16)(9) + (16)(3) +(- 3)(9) +(- 3)(3) (144) +(48) +(- 27) + (- 9). We must expand the terms, :)
Wow multiplying 2/2 in the beginning was a great idea to manipulate denominators of those fractions in the bracket! Keep up the good work my friend!
Excellent!
Glad to hear that!
Thank you for your feedback! Cheers!
You are awesome my dear friend PK 😀
You are pinned!
It's knowing when to do that which separates a good mathematician from a plodder... until the plodder has seen so many various series demonstrated that they see the trick and become a good mathematician.
@@PreMath Thanks my man! You are an inspiration my friend! Always glad to watch your videos! Haha
Another way to evaluate A and B in the PFD is by equating coefficients.
We have A(x + 1) + Bx ≡ 1
⇒ (A + B)x + A ≡ 0x + 1
∴ A + B = 0 equating coefficients in x
and A = 1 equating the constant terms
∴ B = -1.
Yes that's how I solve this kind og problem
I do know his method but I use it in Complex world
I studieded partial fraction decomposition during my first year at the University (degree in Mathematics), many years ago. I taugh this topic for about 40 years.
It is a pleasure mixed with nostalgia to see it reproposed flawlessy.
Love and prayers from Italy!
This is a famous method of evaluation being called as Vn method/telescopic series here in India.
Cheers, Am addicted to your questions!
Thankyu!
Excellent!
Glad to hear that!
Thank you for your feedback! Cheers!
You are awesome 😀
Love and prayers from the USA!
Good Problem !!!!
Please keep uploading these kind of good and tough and challenging problems!!!😀
Today I learnt the concept of telescopic series. Refreshed the partial fraction decomposition. Thank you.
Thanku for providinh us a excelent channel for the lover of mathematics😘
You are very welcome.
Thank you for your feedback! Cheers!
You are awesome Subham 😀
Love and prayers from the USA!
Yea, a really beautiful trick! Thank you so much once again.
PFD excellent solution, Thank you teacher 🙏.
You are very welcome.
Thank you for your feedback! Cheers!
You are awesome 😀
One can also look at the numbers 28, 36, 45, 55, 66, 78,... and notice a pattern that the difference between 2 numbers are in AP with common difference 1. So we can find a formula that generates these numbers by telescopic sum of the difference of the terms Tn - Tn-1, then find a formula for Tn= (1/2)*(n+6)*(n+7). Now if we apply this here, we will get another sum i = 1 to 93 of (2/((i+6)*(i+7))), that gives 2*(1/7 -1/100) = 9/350.
My apologies. 93/350.
7:49 Can someone please explain why you can set it equal to zero? Wouldn't this mean that you divide 1 by zero on the top-left, which is an undefined move?
Please look at the following link about cover up method ruclips.net/video/fgPviiv_oZs/видео.html
1/(x(x+1)=A/X+B/(X+1) USE COVER UP METHOD A=1/(0+1)=1 B=1/(-1)=-1
Application of partial fracture does not occur to a person like me.
Excellent working.
Thanks for the video, the 2/2 trick was very creative.
I'm a bit confused - in partial fraction decomposition, if x(x+1) = 0, then the fraction 1/((x)(x+1)) is undefined? - dividing by zero. Maybe I am not seeing this correctly.
Consider it this way instead: when you have the equation 1= A(x+1) +Bx, this will be true for all values of x, so you can substitute any value any for x that you want to solve for A and B. The easiest thing to do here is set x to some value so that either A or B is removed from the equation. That would be x=0 (to remove B) or x=-1 (to remove A).
You are very welcome Johnny.
You are absolutely correct, denominator must not be zero!
However, we used this idea in the PFD!
Thank you for your feedback! Cheers!
You are awesome 😀
At the point where you set x(x+1)=0 to find A and B, you've already multiplied both sides by x(x+1) and made the denominators go away. So there isn't a problem.
@@Ensign_Cthulhu Yes but in doing so you can get false solutions so you have to be careful with it
if you assume x(x+1)=0, won't that make the initial fraction as division by 0 situation??
I'll use notebook and pen for your videos.We used to solve HOT questions when we were in 10th. you reminded me the memories of that. I'll use notebook for your videos afterwards
Excellent!
Glad to hear that!
Thank you for your feedback! Cheers!
You are awesome Akshay dear 😀
Love and prayers from the USA!
you explained the solution to this number series very well, thanks for sharing
Now I understood why it is very important to know that:
1/(x(x+1)) = (1/x) - (1/(x+1))
Thank you very successful mathematician and youtuber!!😃
Even one cannot calculate using a calculator unless he/she develops each term of the series!
Thank you for a challenging question which you solved quite nicely.
How can we exactly predict the cutting method at +.......+?
Solve a problem, equation. Calculate, determine an integral, the value of something.
Please give the solution of type
X-sinX=A (in this equation X is unknown variable, A is any real number, and angles are measure in radians).
Very informative!
Thank you🙏🙏🌹
You are very welcome.
Thank you for your feedback! Cheers!
You are awesome Engineer 😀
Love and prayers from the USA!
Thank you for your great videos...
And I want to ask you:
I found from wolfram alpha that
sum of k^(-k) from k=1 to k=infinity is: 1.29129 but why?
You are very welcome Mahmoud.
Thank you for your feedback! Cheers!
You are awesome 😀
Because greater the k, smaller the 1/k^k is. The sum would look like this 0 + 1/1 + 1/4 + 1/9... And more of them you add up, closer the sum is to 1.292929... but never passes that point.
Very good.
Maybe define the serie first ?
I dont see any evident pattern to go from 28 to 36 to 45
Obviously the result depends on what ... represents. If it is 0 then you just add 4 items together and voila ;)
Thank you for your video
Excellent
So nice of you.
Thank you for your feedback! Cheers!
You are awesome John 😀
Can you send the solution equation of type x-sinx=a
Nice video
Thank you for your feedback! Cheers!
You are awesome 😀
Thnx a lot
You are very welcome.
Thank you for your feedback! Cheers!
You are awesome Pranav 😀
Love and prayers from the USA!
Pleasr give the solution of equation of type x-sinx=a
Dear Pushpa, please explain what exactly you are asking for?
Thank you
this is a equation x-sinx=a (in this x is unknown variable ,a is any real number and angles are measure in radians)
Nice
Thank you for your feedback! Cheers!
You are awesome Susen 😀
This time the last term remains :D
Thank you for your feedback! Cheers!
You are awesome Anmol😀
Unable to believe that i am able to do these questions xD ...
wish knew these stuffs last year .. now i am 12th passout and dropper :(
(class 13student)
Very nice problem
Excellent!
Glad to hear that!
Thank you for your feedback! Cheers!
You are awesome Mahalakshmi 😀
Love and prayers from the USA!
thankyou Sir
You are very welcome.
Thank you for your feedback! Cheers!
You are awesome Usman 😀
good trick
π/2=log(i)/i , where i=√-1 prove this .This is my challenge 🤪
It would actually be 93/1400 !!!😊
First
Excellent!
You are awesome COS😀
-x=2(2x) no❌.-
x=1÷2(2x) yes✔.
-not-
-not-
-not-
-not-
-not-
Problem no 216
I have a problem and Im wondering if anyone can help me prove it
So 12 x 12 is 144
16 x 9 is 144
but 16 x 9 = 13 x 12
And 13 x 12 is 156
Very easy but
How?
See that (16-3)(9+3) = (13)(12)
but
(16- 3)(9+3) is NOT = (16)(9)
It is
(16)(9) + (16)(3) +(- 3)(9) +(- 3)(3)
(144) +(48) +(- 27) + (- 9).
We must expand the terms, :)
@@albertaraujo6304 oh god that's a lot of math right there
Thanks
Super method, but I have failed.
No worries
Thank you for your feedback! Cheers!
You are awesome Mustafiz 😀
not easy to solve even with calculators because it is hard to put in the series
You should learn programming xD
عاااااش
Stay blessed Omar😀
So simple, the answer is 186/700. Almost ten minutes, for now, watch that long video is not my project.