This is a nice exercise in integrating a rational function. For me, the surprise was the way you factorised the quartic x⁴-x²+1: x⁴-x²+1=x⁴+2x²+1-3x²=(x²+1)²-3x² =(x²+1+√3x)(x²+1-√3x) =(x²+√3x+1)(x²-√3x+1) How did you come up with this "trick"? Here is my method (without using your trick) which is rather longer. We note that x⁶+1=(x²+1)(x⁴-x²+1). (This comes from the standard factorisation x³+1=(x+1)(x²-x+1), replacing x by x²) On the other hand, x⁶+1=0 ⇔x⁶=-1 which, as -1=cis(π), has solutions x=cis(±π/6), cis(±3π/6), cis(±5π/6). cis(±3π/6)=cis(±π/2)=±i leading to factors x-i, x+i with product (x-i)(x+i)=x²+1. So x=cis(±π/6), cis(±5π/6) are the solutions of x⁴-x²+1=0. The roots cis(±π/6) have sum 2cos(π/6)=√3 and product 1, so satisfy the quadratic equation x²-√3x+1=0. The roots cis(±5π/6) have sum 2cos(5π/6)=-2cos(π/6)=-√3 and product 1, so satisfy the quadratic equation x²+√3x+1=0. Hence x⁴-x²+1=(x²+√3x+1)(x²-√3x+1). Now, once I know this, I can write (x²+√3x+1)(x²-√3x+1) =(x²+1+√3x)(x²+1-√3x)=(x²+1)²-3x² =x⁴+2x²+1-3x²=x⁴-x²+1 Now I can run the argument backwards to get what you wrote! So my question is: did you also follow this route to arrive at your "trick", or is there some easy way to reach it?
That is not a "trick" it is one of the general methods used to factorize bi quadratics. We will complete the square and then use difference of squares. If you are interested I've got a algebra problem based on this technique Find summation r going from 1 to 20 r/(r⁴ + 1/4)
@@ashvanthvijai7220 I've checked my notes on solving quartics, and was reminded of Ferrari's method: The aim is to factorise the quartic (with real coefficients) into the product of two quadratics (with real coefficients), which is possible by the conjugate root theorem. To do this, we rewrite the quartic equation as the square of a quadratic equals the square of a linear expression. (Or, equivalently, rewrite the quartic expression as the square of a quadratic minus the square of a linear expression). Start by rewriting the quartic as x⁴=x²-1 Since the quartic is monic and depressed (i.e. has no x³ term), our quadratic must be of the form x²+m. Since (x²+m)²=x⁴+2mx²+m², adding 2mx²+m² to each side of our quartic equation we get the equivalent equation (x²+m)²=2mx²+m²+x²-1 or (x²+m)²=(2m+1)x²+m²-1 The RHS will be a perfect square iff m²=1, and as we want 2m+1≥0 (to get the square of a linear expression with real coefficients) we take m=1: (x²+1)²=3x² (x²+1)²-3x²=0 which is what was obtained in the video. Is that what you had in mind? When you said "a general method to factorise biquadratics" I'm not sure whether you were referring to a technique for quartics in general or for the easy case of quadratics in x² which appears in the video. I look forward to having a go at the problem you suggested.
@@learn-with-vince Lol! Algebra, algebra, algebra! But I took on board @ashvanthvijai7220's suggestion about using the difference of two squares to factorise quartics, as you can see from my reply above.
To see me check my work take a look at this derivative: th-cam.com/video/KaY2grN7HYk/w-d-xo.html
You forgot to add +c
You're right ! That's on me.
😅
Ok now derivate the answer. Lets see it turn back into 1/(x^6 +1)
Here you go: th-cam.com/video/KaY2grN7HYk/w-d-xo.html
@@learn-with-vince Thank you, I actually wanted to see that monster return to its humble beginnings
Ah, so thats how it is. Hmm. Ah.
Definitely a long problem.
love the quick format also love "lawn" 😅
Glad you enjoy it!
Great format !
had to retake a math course and ill probably just use this as my study guide. thanks vince.
Glad I could help!
To think a derivative and an integral are related but one is extremely hard compared to the other
FTC!
“+C”: am I a joke to you
Yes... no but really I did just forget it that's on me.
Why not use series?
lawn
LAWN
@@learn-with-vince mower
can we use partial fraction with complex root?
I don't see why not
@@learn-with-vince thnks this will make it hell of much easier
Where is the c? Zero point
i am daed
It's a tough one
This is a nice exercise in integrating a rational function.
For me, the surprise was the way you factorised the quartic x⁴-x²+1:
x⁴-x²+1=x⁴+2x²+1-3x²=(x²+1)²-3x²
=(x²+1+√3x)(x²+1-√3x)
=(x²+√3x+1)(x²-√3x+1)
How did you come up with this "trick"?
Here is my method (without using your trick) which is rather longer.
We note that x⁶+1=(x²+1)(x⁴-x²+1).
(This comes from the standard factorisation x³+1=(x+1)(x²-x+1), replacing x by x²)
On the other hand,
x⁶+1=0 ⇔x⁶=-1
which, as -1=cis(π), has solutions x=cis(±π/6), cis(±3π/6), cis(±5π/6).
cis(±3π/6)=cis(±π/2)=±i leading to factors x-i, x+i with product (x-i)(x+i)=x²+1.
So x=cis(±π/6), cis(±5π/6) are the solutions of x⁴-x²+1=0.
The roots cis(±π/6) have sum 2cos(π/6)=√3 and product 1, so satisfy the quadratic equation x²-√3x+1=0.
The roots cis(±5π/6) have sum 2cos(5π/6)=-2cos(π/6)=-√3 and product 1, so satisfy the quadratic equation x²+√3x+1=0.
Hence x⁴-x²+1=(x²+√3x+1)(x²-√3x+1).
Now, once I know this, I can write
(x²+√3x+1)(x²-√3x+1)
=(x²+1+√3x)(x²+1-√3x)=(x²+1)²-3x²
=x⁴+2x²+1-3x²=x⁴-x²+1
Now I can run the argument backwards to get what you wrote!
So my question is: did you also follow this route to arrive at your "trick", or is there some easy way to reach it?
That is not a "trick" it is one of the general methods used to factorize bi quadratics. We will complete the square and then use difference of squares. If you are interested I've got a algebra problem based on this technique
Find summation r going from 1 to 20 r/(r⁴ + 1/4)
I actually did do this using Rothwell's longer way
@@ashvanthvijai7220 I've checked my notes on solving quartics, and was reminded of Ferrari's method:
The aim is to factorise the quartic (with real coefficients) into the product of two quadratics (with real coefficients), which is possible by the conjugate root theorem.
To do this, we rewrite the quartic equation as the square of a quadratic equals the square of a linear expression.
(Or, equivalently, rewrite the quartic expression as the square of a quadratic minus the square of a linear expression).
Start by rewriting the quartic as
x⁴=x²-1
Since the quartic is monic and depressed (i.e. has no x³ term), our quadratic must be of the form x²+m.
Since (x²+m)²=x⁴+2mx²+m²,
adding 2mx²+m² to each side of our quartic equation we get the equivalent equation
(x²+m)²=2mx²+m²+x²-1
or (x²+m)²=(2m+1)x²+m²-1
The RHS will be a perfect square iff m²=1, and as we want 2m+1≥0 (to get the square of a linear expression with real coefficients) we take m=1:
(x²+1)²=3x²
(x²+1)²-3x²=0
which is what was obtained in the video.
Is that what you had in mind? When you said "a general method to factorise biquadratics" I'm not sure whether you were referring to a technique for quartics in general or for the easy case of quadratics in x² which appears in the video.
I look forward to having a go at the problem you suggested.
@@learn-with-vince Lol! Algebra, algebra, algebra!
But I took on board @ashvanthvijai7220's suggestion about using the difference of two squares to factorise quartics, as you can see from my reply above.
@@MichaelRothwell1 yeah it can't be used to factorize every bi quadratic. It can be used to factorize depressed ones.