MedVision ad

Reduction Formula Question (1 Viewer)

hscstudnet

Member
Joined
Jan 16, 2020
Messages
59
Gender
Male
HSC
2020
I am struggling to do this question. I tried solving it and yeah thats pretty much it. I tried :(.
Bored.JPG
Also generally, I can do integration by parts questions very well but whenever it comes to reduction formula or proof by Induction questions where there is a recursive formula involved, I only get it right 30-50% of the times. What are some clues or hints I should look for when proving the formula in both these types of questions.

Thanks in advance.
 

quickoats

Well-Known Member
Joined
Oct 26, 2017
Messages
970
Gender
Undisclosed
HSC
2019
The hint is in the “algebraic manipulation” part of the question. Is there anything you can do to the numerator to help you get the required result?

With reduction questions, you kinda just have to trial and error a solution through recognition. As you do more questions, you’ll be able to recognise bits (from integration by parts) that pop up every now and then.
 

hscstudnet

Member
Joined
Jan 16, 2020
Messages
59
Gender
Male
HSC
2020
The hint is in the “algebraic manipulation” part of the question. Is there anything you can do to the numerator to help you get the required result?

With reduction questions, you kinda just have to trial and error a solution through recognition. As you do more questions, you’ll be able to recognise bits (from integration by parts) that pop up every now and then.
Thank you so much I got the answer
 

CM_Tutor

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,642
Gender
Male
HSC
N/A
My first approach would be to rearrange the integrand to facilitate integration - a change like . Having an and an in the formula gives a clue as to what to separate.

In this case, this would give:

 

CM_Tutor

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,642
Gender
Male
HSC
N/A
Another approach is to rearrange the target result, in this case by examining :



This approach removes the need to determine what to separate (as in the above approach) but is difficult to apply if the coefficient of the term in or is not .
 

hscstudnet

Member
Joined
Jan 16, 2020
Messages
59
Gender
Male
HSC
2020
Another approach is to rearrange the target result, in this case by examining :



This approach removes the need to determine what to separate (as in the above approach) but is difficult to apply if the coefficient of the term in or is not .
Thanks for both the methods do you have any tips on proof questions also inequality proofs there are pretty annoying
 

CM_Tutor

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,642
Gender
Male
HSC
N/A
Thanks for both the methods do you have any tips on proof questions also inequality proofs there are pretty annoying
Probably best to ask specific questions / examples - I'll put in advice where I can. :)
 
Joined
Mar 7, 2020
Messages
10
Gender
Undisclosed
HSC
N/A
I am struggling to do this question. I tried solving it and yeah thats pretty much it. I tried :(.
View attachment 28593
Also generally, I can do integration by parts questions very well but whenever it comes to reduction formula or proof by Induction questions where there is a recursive formula involved, I only get it right 30-50% of the times. What are some clues or hints I should look for when proving the formula in both these types of questions.

Thanks in advance.
thats a rare type of reduction question. The vast majority use IBP or a trig identity.
 

Trebla

Administrator
Administrator
Joined
Feb 16, 2005
Messages
8,391
Gender
Male
HSC
2006
As a rule of thumb, if you see reduction formulae in the form:

and if b is independent of n then it is likely that a simple addition approach would work.

Integration by parts usually generates a recurrence of the original integral as well as a bunch of factors dependent on n (due to the nature of differentiating and integrating involved).
 

hscstudnet

Member
Joined
Jan 16, 2020
Messages
59
Gender
Male
HSC
2020
As a rule of thumb, if you see reduction formulae in the form:

and if b is independent of n then it is likely that a simple addition approach would work.

Integration by parts usually generates a recurrence of the original integral as well as a bunch of factors dependent on n (due to the nature of differentiating and integrating involved).
🤯🤯 This is a such good way to identify the type, thank you so much.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top