Some four unit question (1 Viewer)

[exiting]

Not sure if you can visualise or if you need any visualising.

Consider the area under the curve y = 1/x bounded by the x-axis between x = n, and x = n+1 (for n is greater than or equal to 1).

(a) By taking the area of the upper rectangle and the lower rectangle, show that
1/(n+1) < ln (1+1/n) < 1/n

(b) Hence, deduce that
(1+1/n)^n < e < (1+1/n)^(n+1)

This was some 4 unit question that was on my test in the half yearly, last question may I add! Thanks in advance.

 

[Drongoski]

 

[alakazimmy]

For part a, compare the areas under the graph. The upper rectangle (1/n) is greater than the actual area (integrate), and this area is greater than the lower rectangle (1/(n+1))

For part b, use the inequalities from part a.

Using just the inequality on the LHS:
We have 1/(n+1) < ln (1 + 1/n)

1< (n+1) ln (1 + 1/n). Raising both sides to the power of e, we obtain:
e < (1 + 1/n)^(n+1)

using the inequality on the RHS:
ln (1 + 1/n) < 1/n
n ln (1 + 1/n) < 1. Again, raise both sides to the power of e to obtain:
(1 + 1/n)^n < e

Hence, (1 + 1/n)^n < e < (1 + 1/n)^(n+1)

 

[alakazimmy]

Beat me to it

 

[Drongoski]

alakazimmy


I'm so sorry; you merit equal credit.

 

[exiting]

Did you guys/girls, know how to do this in year 12? :| Thanks very much! Glad everyone dropped 2 marks

 

[Trebla]

Note that as n approaches infinity we derive the famous result:



A similar question actually appeared in a past HSC Extension 1 paper...

 

[shaon0]

Not sure if you can visualise or if you need any visualising.

Consider the area under the curve y = 1/x bounded by the x-axis between x = n, and x = n+1 (for n is greater than or equal to 1).

(a) By taking the area of the upper rectangle and the lower rectangle, show that
1/(n+1) < ln (1+1/n) < 1/n

(b) Hence, deduce that
(1+1/n)^n < e < (1+1/n)^(n+1)

This was some 4 unit question that was on my test in the half yearly, last question may I add! Thanks in advance.

i've seen this in a 3unit test.

 

[L]

brb mate,

coles

 

[alakazimmy]

alakazimmy


I'm so sorry; you merit equal credit.

No worries, it's all good

 

[exiting]

Note that as n approaches infinity we derive the famous result:



A similar question actually appeared in a past HSC Extension 1 paper...

I think I read in Cambridge somewhere about that e result, but how does umm that results turn into e = 1/0! + 1/1! + 1/2! + ... 1/n! ?

 

[Drongoski]

You learn this under the Maclaurin Series expansion for e^x and set x = 1

 

[exiting]

What a turn off from Uni... LOL. Thanks again

 

[alakazimmy]

Haha, there's like a million different definitions for e.

 

[Trebla]

I think I read in Cambridge somewhere about that e result, but how does umm that results turn into e = 1/0! + 1/1! + 1/2! + ... 1/n! ?

Expand (1 + 1/n)ⁿ using binomial expansions (simplifying all factorial notations) and take the limit as n approaches infinity, then you get that result.