Mathematical Induction Inequality (1 Viewer)

[Eazi]

wait until you get to exercise 3

 

[Average Boreduser]

wait until you get to exercise 3

im assuming u mean the trig one w x^2+1/x^2. Hint for anyone doing MI rn: you need to use strong induction (ie prove 2 cases) to prove by induction- and also recall triple angle formula

 

[Average Boreduser]

i remember some of the geometry ones (cancerous)

 

[Eazi]

im assuming u mean the trig one w x^2+1/x^2. Hint for anyone doing MI rn: you need to use strong induction (ie prove 2 cases) to prove by induction

I finished the whole book (assuming you're in a1) I just can't do the last ex3 lv3 question (independent one)

 

[Average Boreduser]

I finished the whole book (assuming you're in a1) I just can't do the last ex3 lv3 question (independent one)

the one w T1=kjfhkj and the sequence?

 

[Eazi]

yeah like the one to show that Tn=(sqrroot2)^n+2 etc

 

[Average Boreduser]

yeah like the one to show that Tn=(sqrroot2)^n+2 etc

its a strong induction. just brute force it. I think its supposed to be a trek

 

[Average Boreduser]

yeah like the one to show that Tn=(sqrroot2)^n+2 etc

Here I'll quickly list out your steps:
1. Prove true for n=1
2. create ur assumption, to prove true for n=k+1
- use the identity you were provided in the qn i.e. tn= 2tn-1 ..... etc
take this as ur lhs of (2) and ur rhs as the original sqrtcospi/4 (k+1) wtv that you subbed as n=k+1. - To use this though, you need to prove again (ie strong induction) for another integer (i'll let u figure that out)
look at lhs of (2), use compound angle and you should get something with like sqrt(2)^smn [cos(kpi/4)-sinkpi/4]
then observe the rhs of (2), expand cmpnd angle, you get same as lhs.

 

[liamkk112]

I finished the whole book (assuming you're in a1) I just can't do the last ex3 lv3 question (independent one)

send the question if u want

 

[Average Boreduser]

send the question if u want

 

[liamkk112]

View attachment 42239

ill skip base cases as its pretty self explanatory

assumption:
and we are assuming for n = k, k>= 3 (we have already proved n = 1, n= 2 in the base cases so we will also assume these are true)

proving n = k + 1:
RTP
LHS (using the sequence definition)


by the cos angle difference formula, where we are taking the difference of kpi/4 and -pi/4
because cos, sin evaluated at pi/4 = sqrt2 /2
by cos angle sum of kpi/4 and pi/4
then because we get that as required

hence by mathematical induction ...

 

[Luukas.2]

1. Using Mathematical induction, prove that n^2 + 1 > n for all integers n>1
For n=1
LHS= (2)^2+1= 5
RHS=2

Thus LHS>RHS, therefore true

Assume n=k,
k^2+1 >K

RTP; n=k+1
(k+1)^2 +1 > k+1

Now what do I do and how do I go about it

If you are still around, one approach with inequalities are to consider LHs > RHS in the form of LHS - RHS... So, in this case:

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This can make it easier to see how to make use of the induction hypothesis, at times.