if someone could explain how to use the coverup rule for 4u partial frac would be appreciated (1 Viewer)

[ekjchale#1]

tried to do a few questions but couldnt get the hang of it, and ended up being way longer than how i would do it normally
thanks in advance

 

[5uckerberg]

Okay so the idea for the well-known Heaviside cover-up method goes like this

Suppose you have something like

What happens is that you split this into partial fractions which will give us . Intuitively you would cover one of the factors or with your finger and then say oh let x=-3 or let x=.

At this point, you might be wondering why we make such statements. First and foremost it will allow us to find the values for A and B and also the backbone of the Heaviside cover-up method goes like this

From we multiply by or on both sides so that we get a constant which is either A or B and if you think about it a constant is your best friend for integration because they will stay the same regardless of the value you have for x.

To show my proof what is actually happening is that from multiplying by will give us . Do you see the constant A, well now we will say let x=-3 because you want for the fraction because 0 divided by anything on the denominator is still 0. As a result of this our working will be easy because just sub in x=-3 on the LHS and you should get . Repeat this process and you will have .

Now you you remember when I said that we can do this over here "Intuitively you would cover one of the factors or with your finger and then say oh let or let ." The reason why we could do that is because of the backbone of the Heaviside cover-up method as I explained earlier from the previous two paragraph.

I hope this helps you understand how the cover-up rule works and also why it works so well. If you still have any difficulties feel free to reply on this post and someone here will explain it more concisely for you.

 

[idkkdi]

tried to do a few questions but couldnt get the hang of it, and ended up being way longer than how i would do it normally
thanks in advance

coverup rule is really just doing what you normally do but skipping more steps. might want to better understand how it comes about.

 

[ekjchale#1]

Okay so the idea for the well-known Heaviside cover-up method goes like this

Suppose you have something like

What happens is that you split this into partial fractions which will give us . Intuitively you would cover one of the factors or with your finger and then say oh let x=-3 or let x=.

At this point, you might be wondering why we make such statements. First and foremost it will allow us to find the values for A and B and also the backbone of the Heaviside cover-up method goes like this

From we multiply by or on both sides so that we get a constant which is either A or B and if you think about it a constant is your best friend for integration because they will stay the same regardless of the value you have for x.

To show my proof what is actually happening is that from multiplying by will give us . Do you see the constant A, well now we will say let x=-3 because you want for the fraction because 0 divided by anything on the denominator is still 0. As a result of this our working will be easy because just sub in x=-3 on the LHS and you should get . Repeat this process and you will have .

Now you you remember when I said that we can do this over here "Intuitively you would cover one of the factors or with your finger and then say oh let or let ." The reason why we could do that is because of the backbone of the Heaviside cover-up method as I explained earlier from the previous two paragraph.

I hope this helps you understand how the cover-up rule works and also why it works so well. If you still have any difficulties feel free to reply on this post and someone here will explain it more concisely for you.

Okay so the idea for the well-known Heaviside cover-up method goes like this

Suppose you have something like

What happens is that you split this into partial fractions which will give us . Intuitively you would cover one of the factors or with your finger and then say oh let x=-3 or let x=.

At this point, you might be wondering why we make such statements. First and foremost it will allow us to find the values for A and B and also the backbone of the Heaviside cover-up method goes like this

From we multiply by or on both sides so that we get a constant which is either A or B and if you think about it a constant is your best friend for integration because they will stay the same regardless of the value you have for x.

To show my proof what is actually happening is that from multiplying by will give us . Do you see the constant A, well now we will say let x=-3 because you want for the fraction because 0 divided by anything on the denominator is still 0. As a result of this our working will be easy because just sub in x=-3 on the LHS and you should get . Repeat this process and you will have .

Now you you remember when I said that we can do this over here "Intuitively you would cover one of the factors or with your finger and then say oh let or let ." The reason why we could do that is because of the backbone of the Heaviside cover-up method as I explained earlier from the previous two paragraph.

I hope this helps you understand how the cover-up rule works and also why it works so well. If you still have any difficulties feel free to reply on this post and someone here will explain it more concisely for you.

Aye thanks bro, are there any situations or redflags which tell us not to use it?

 

[mmmmmmmmaaaaaaa]

Whenever there is more than one unknown, you need to use simultaneous equations or some other method.

I would always try to use the cover-up method first - a time saver

 

[Run hard@thehsc]

coverup rule is really just doing what you normally do but skipping more steps. might want to better understand how it comes about.

fair - is it better to skip to the normal route if this is kinda not familiar?

 

[stupid_girl]

Aye thanks bro, are there any situations or redflags which tell us not to use it?

Cover up method only works for a linear factor in its highest degree.

Consider 1/[(x-1) (x-2)^2 (x-3)^3 (x^2+x+1)] = A/(x-1) + B/(x-2) + C/(x-2)^2 + D/(x-3) + E/(x-3)^2 + F/(x-3)^3 + (Gx+H)/(x^2+x+1).
It only works for A, C and F.

 

[dan964]

tried to do a few questions but couldnt get the hang of it, and ended up being way longer than how i would do it normally
thanks in advance

The cover up method effectively you are treating x as a parameter to find A and B.


Basically if you times both sides by the denominator you get:

So you can pick such an x so that 2x+1 = 0 (x=1/2), and another x such that x+3=0 (x=-3)
This gives you two very easy expressions for A and B since the terms cancel out due to multiplication by 0.

The cover up method basically is a way of condensing these first 2 steps into 1 step. (You are basically doing that expansion calculation in your head)

The reason why you do this is to split a complex fraction into two smaller fractions which are easier for integration

I would only recommend cover up method for simpler examples. More complex, like the example in the above comment, best to expand out.

 

[idkkdi]

fair - is it better to skip to the normal route if this is kinda not familiar?

ye dont bother with coverup rule.

 

[stupid_girl]

The cover up method effectively you are treating x as a parameter to find A and B.


Basically if you times both sides by the denominator you get:

So you can pick such an x so that 2x+1 = 0 (x=1/2), and another x such that x+3=0 (x=-3)
This gives you two very easy expressions for A and B since the terms cancel out due to multiplication by 0.

The cover up method basically is a way of condensing these first 2 steps into 1 step. (You are basically doing that expansion calculation in your head)

The reason why you do this is to split a complex fraction into two smaller fractions which are easier for integration

I would only recommend cover up method for simpler examples. More complex, like the example in the above comment, best to expand out.

Even in more complex examples, it is still advisable to use the cover up method as the first step to reduce the number of unknowns.

Example 1
1/[(x+1) (x-1) (x-2)^2] = A/(x+1) + B/(x-1) + C/(x-2) + D/(x-2)^2
By cover up method,
A=1/[(-1-1) (-1-2)^2]=-1/18
B=1/[(1+1) (1-2)^2]=1/2
D=1/[(2+1) (2-1)]=1/3
Now, it's much easier to deal with C.
In this case, it is convenient to substitute x=0.
1/[(0+1) (0-1) (0-2)^2] = (-1/18)/(0+1) + (1/2)/(0-1) + C/(0-2) + (1/3)/(0-2)^2
-1/4 = -1/18 - 1/2 - C/2 + 1/12
C=-4/9

Example 2
1/[(x-1) (x-2) (x^2+x+1)] = A/(x-1) + B/(x-2) + (Cx+D)/(x^2+x+1)
By cover up method,
A=1/[(1-2)(1^2+1+1)]=-1/3
B=1/[(2-1)(2^2+2+1)]=1/7
Now, there are only 2 unknowns left.
Substitute x=0.
1/[(0-1) (0-2) (0^2+0+1)] = (-1/3)/(0-1) + (1/7)/(0-2) + (C(0)+D)/(0^2+0+1)
1/2 = 1/3 - 1/14 + D
D=5/21
Substitute x=-1.
1/[(-1-1) (-1-2) ((-1)^2-1+1)] = (-1/3)/(-1-1) + (1/7)/(-1-2) + (C(-1)+(5/21))/((-1)^2-1+1)
1/6 = 1/6 - 1/21 - C + 5/21
C=4/21

 

[dan964]

Even in more complex examples, it is still advisable to use the cover up method as the first step to reduce the number of unknowns.

Example 1
1/[(x+1) (x-1) (x-2)^2] = A/(x+1) + B/(x-1) + C/(x-2) + D/(x-2)^2
By cover up method,
A=1/[(-1-1) (-1-2)^2]=-1/18
B=1/[(1+1) (1-2)^2]=1/2
D=1/[(2+1) (2-1)]=1/3
Now, it's much easier to deal with C.
In this case, it is convenient to substitute x=0.
1/[(0+1) (0-1) (0-2)^2] = (-1/18)/(0+1) + (1/2)/(0-1) + C/(0-2) + (1/3)/(0-2)^2
-1/4 = -1/18 - 1/2 - C/2 + 1/12
C=-4/9

Example 2
1/[(x-1) (x-2) (x^2+x+1)] = A/(x-1) + B/(x-2) + (Cx+D)/(x^2+x+1)
By cover up method,
A=1/[(1-2)(1^2+1+1)]=-1/3
B=1/[(2-1)(2^2+2+1)]=1/7
Now, there are only 2 unknowns left.
Substitute x=0.
1/[(0-1) (0-2) (0^2+0+1)] = (-1/3)/(0-1) + (1/7)/(0-2) + (C(0)+D)/(0^2+0+1)
1/2 = 1/3 - 1/14 + D
D=5/21
Substitute x=-1.
1/[(-1-1) (-1-2) ((-1)^2-1+1)] = (-1/3)/(-1-1) + (1/7)/(-1-2) + (C(-1)+(5/21))/((-1)^2-1+1)
1/6 = 1/6 - 1/21 - C + 5/21
C=4/21

Agreed but mainly for linear factors to avoid likelihood of error. But you are probably a pro at it so disregard haha.
Otherwise multiplying through does the same trick and only is an extra line of working.