What is the value of e when substituting n=infinity into the limit (1 + 1/n)^n?

[tahayassen]

Homework Statement

Calculate the value of e.

Homework Equations

The Attempt at a Solution

If I substitute n=infinity into that limit, wouldn't I get e = 1?

(1 + 1/(infinity))^infinity

As the denominator of 1/infinity approaches very large values, 1/infinity approaches zero.

Therefore, (1 + 0)^infinity

e = 1

 

[SammyS]

Homework Statement

Calculate the value of e.

Homework Equations

The Attempt at a Solution

If I substitute n=infinity into that limit, wouldn't I get e = 1?

(1 + 1/(infinity))^infinity

As the denominator of 1/infinity approaches very large values, 1/infinity approaches zero.

Therefore, (1 + 0)^infinity

e = 1

No, you won't get 1 .

Try some successively larger values for n to see what happens !

 

[tahayassen]

No, you won't get 1 .

Try some successively larger values for n to see what happens !

Ya, but I don't understand why.

 

[DryRun]

Not so long ago, i did the same mistake as you did in post #1.

Basically, you're trying to find the limit of: \lim_{n\to \infty}n^n
Let\; y=\left( 1+ \frac{1}{n} \right)^n Hint: Take logs on both sides.

 

[Steely Dan]

Ya, but I don't understand why.

Try writing out the polynomial for a given n:
s_n = \left(\frac{1}{n}\right)^n (1+n)...(1+n) = \left(\frac{1}{n}\right)^n (1 + a_1 n + a_2 n^2 + ... + a_{n-1} n^{n-1} + a_n n^n)<br />
where the coefficients are simply those of Pascal's triangle. Obviously as n\rightarrow \infty, the former terms are small compared to the latter, so
s_n \approx \frac{1}{n^n} ((n-1) n^{n-1} + n^n),
since the second to last term in a row of Pascal's triangle is always the number of the row minus one. Therefore
s_n \approx \frac{1}{n^n} (2 n^n - n^{n-1}) = 2 - \frac{1}{n}
Obviously the latter term drops off as n gets large, but the contribution it gave to the term of order n^0 sticks behind. Try it with keeping the contribution to that order with three terms, four, etc.

Edit: And I don't think this is a dumb question. I had to come up with that reasoning when I just read your post. I suppose it serves to illuminate that sometimes our intuition about the behavior of large numbers can lead us astray!

 

[tahayassen]

Not so long ago, i did the same mistake as you did in post #1.

Basically, you're trying to find the limit of: \lim_{n\to \infty}n^n
Let\; y=\left( 1+ \frac{1}{n} \right)^n Hint: Take logs on both sides.

I took the log of both sides, and then I moved the n exponent to the coefficient, so I got:

logy = nlog(1+1/n)

But I'm not sure where to proceed from here.

Try writing out the polynomial for a given n:
s_n = \left(\frac{1}{n}\right)^n (1+n)...(1+n) = \left(\frac{1}{n}\right)^n (1 + a_1 n + a_2 n^2 + ... + a_{n-1} n^{n-1} + a_n n^n)<br />
where the coefficients are simply those of Pascal's triangle. Obviously as n\rightarrow \infty, the former terms are small compared to the latter, so
s_n \approx \frac{1}{n^n} ((n-1) n^{n-1} + n^n),
since the second to last term in a row of Pascal's triangle is always the number of the row minus one. Therefore
s_n \approx \frac{1}{n^n} (2 n^n - n^{n-1}) = 2 - \frac{1}{n}
Obviously the latter term drops off as n gets large, but the contribution it gave to the term of order n^0 sticks behind. Try it with keeping the contribution to that order with three terms, four, etc.

Edit: And I don't think this is a dumb question. I had to come up with that reasoning when I just read your post. I suppose it serves to illuminate that sometimes our intuition about the behavior of large numbers can lead us astray!

Wow. I'm totally lost. I didn't even understand the very first step you did.

If I understand correctly, Pascal's triangle is:

So where did you get s_n = \left(\frac{1}{n}\right)^n (1+n)...(1+n) from?

 

[Steely Dan]

Wow. I'm totally lost. I didn't even understand the very first step you did.

If I understand correctly, Pascal's triangle is:

So where did you get that from?

The first step I did is to pull out n^{-n} from the polynomial, leaving s_n = n^{-n} (1+n)^n. Pascal's triangle gives you the coefficients of that latter polynomial, whose lowest order term is surely 1 and whose highest order term is surely n^n.

 

[Mark44]

If I substitute n=infinity into that limit, wouldn't I get e = 1?

(1 + 1/(infinity))^infinity

It's not legal to substitute ∞ into an expression and do arithmetic on it.

As the denominator of 1/infinity approaches very large values, 1/infinity approaches zero.

Therefore, (1 + 0)^infinity

This is one of several indeterminate forms, which include [0/0], [±∞/∞], [∞ - ∞], and [1^∞]. These show up in the context of limits, and indicate that some more work needs to be done.

 

[tahayassen]

The first step I did is to pull out n^{-n} from the polynomial, leaving s_n = n^{-n} (1+n)^n. Pascal's triangle gives you the coefficients of that latter polynomial, whose lowest order term is surely 1 and whose highest order term is surely n^n.

y=(1+n^-1)n
=n^-n(1+n)^n

How did you go from the first step to the second step?

It's not legal to substitute ∞ into an expression and do arithmetic on it.


This is one of several indeterminate forms, which include [0/0], [±∞/∞], [∞ - ∞], and [1^∞]. These show up in the context of limits, and indicate that some more work needs to be done.

What if I do:

Let m = 1/n

lim as m approaches 0 (1+m)^(1/m)
= (1+0)^(1/0)
= 1^(1/0)

 

[Mark44]

y=(1+n^-1)n
=n^-n(1+n)^n

How did you go from the first step to the second step?



What if I do:

Let m = 1/n

lim as m approaches 0 (1+m)^(1/m)
= (1+0)^(1/0)
= 1^(1/0)

It's not legal to divide by zero, either.

 

[tahayassen]

It's not legal to divide by zero, either.

So how would you solve this limit without using the approximation method then?

 

[Mark44]

Continue with what you were doing earlier...

logy = nlog(1+1/n)

log(y) should be ln(y).
Rewrite the right side as [ln (1 + 1/n)]/(1/n).
Take the limit of both sides. If necessary, use L'Hopital's Rule.

 

[tahayassen]

I give up.

I'm trying to learn L'Hopital's Rule. Learning Pascal's Triangle was already hard enough. :(

edit: I'll keep trying, but this is going to take me some time.

 

[DryRun]

L'Hopital's Rule isn't too complicated: here is the formal definition
Basically, it relates to fractions and if you put the value of the limit in the numerator and denominator, and if you get 0/0 or ∞/∞, (which are called indeterminate forms) then you will need to use L'Hopital's Rule. The rule simply requires that you differentiate both the numerator and denominator, independently. In some cases, you'll need to repeat the procedure.

 

[tahayassen]

y=(1+n^-1)n
=n^-n(1+n)^n

How did you go from the first step to the second step?



What if I do:

Let m = 1/n

lim as m approaches 0 (1+m)^(1/m)
= (1+0)^(1/0)
= 1^(1/0)

I just noticed at the end of this: not only did I divide by zero, but I still get 1^infinity at the end, which is an indeterminate form. Opps!

Back on track, okay, so I did that rule, but I still ended up with a value of 1! D=

http://i2.lulzimg.com/4cb849a256.png

y\quad =\quad { (1+{ n }^{ -1 }) }^{ n }\\<br /> lny\quad =\quad { ln(1+{ n }^{ -1 }) }^{ n }\\<br /> lny\quad =\quad nln(1+{ n }^{ -1 })\\<br /> \underset { n\rightarrow \infty }{ lim } lny\quad =\quad \underset { n\rightarrow \infty }{ lim } \frac { ln(1+{ n }^{ -1 }) }{ { n }^{ -1 } } \\<br /> \underset { n\rightarrow \infty }{ lim } lny\quad =\quad \underset { n\rightarrow \infty }{ lim } \frac { \frac { 1 }{ 1+{ n }^{ -1 } } ({ -n }^{ -2 }) }{ { -n }^{ -2 } } \\<br /> \underset { n\rightarrow \infty }{ lim } lny\quad =\quad \underset { n\rightarrow \infty }{ lim } \frac { 1 }{ 1+{ n }^{ -1 } } \\<br /> \underset { n\rightarrow \infty }{ lim } lny\quad =\quad \frac { 1 }{ 1+{ \infty }^{ -1 } } \\<br /> \underset { n\rightarrow \infty }{ lim } lny\quad =\quad \frac { 1 }{ 1+0 }

 

[DryRun]

So, you've reached: \lim_{n\to \infty} \ln y = 1
But, \lim_{n\to \infty} \ln y = \ln \lim_{n\to \infty} y
Now, can you find \lim_{n\to \infty} y?

 

[Mark44]

OK, since lim ln(y) = 1, then y = ?

 

[tahayassen]

ln * lim(y) = 1
lim(y) = e^1
= e

But I still haven't found a value. :|

 

[DryRun]

Here, e, represents the exponential value. Use your calculator. What is the value of e^1?

 

[tahayassen]

The original question was:

Calculate the value of e.

Essentially, what we did was:

e = e
= 2... (using the calculator function)

But how did the calculator get that value?

 

[DryRun]

According to my calculator, e = 2.718281828

But how did the calculator get that value?

LOL! Sorry, i can't go into the electronics that pump out that number from your calculator. It happens due to logic gates. e has a standard value, much like \pi.

 

[tahayassen]

According to my calculator, e = 2.718281828


LOL! Sorry, i can't go into the electronics that pump out that number from your calculator.

How do mathematicians know what e is equal to?

How did they calculate it?

Did they just use very large values of n?

 

[DryRun]

e = 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + ...

 

[tahayassen]

Alright. Thank you everyone for your time.

I think I understand it now. I was surprised to find that 1^infinity is an indeterminate form! I thought it would just equal 1.

 

[DryRun]

The historical development of e using various methods: http://www.intmath.com/exponential-logarithmic-functions/calculating-e.php

 

[Steely Dan]

y=(1+n^-1)n
=n^-n(1+n)^n

How did you go from the first step to the second step?

I expanded that polynomial out. Ignoring the bit about Pascal's triangle, it's straightforward to see that the polynomial expands to:
s_n = n^{-n} (1 + (n-1)n + ... + (n-1)n^{n-1} + n^n)
If you don't believe it, try it for a few small values of n to see it. Now, if you do accept that, then what I said afterwards follows. That is, to a first approximation, we can drop everything but the last two terms in the polynomial, which gives e = 2 after we take the limit. If we keep instead the first three terms, we have
s_n \approx n^{-n} \left(\frac{1}{2}(n^2 -3n +2)n^{n-2} + (n-1)n^{n-1} + n^n\right)
Trying the same strategy, in that first term there is only one term with order n^n, so
s_n \approx n^{-n} (\frac{1}{2} n^n + n^n + n^n) = 2.5.
So the idea is that the more terms you keep, the closer you get to the actual value of e. I think this leads to the Taylor expansion for the exponential function directly, but I haven't worked it out.

By the way -- the "argument" using the natural logarithm is not much of an argument, because you're assuming the nature of e to prove the nature of e. It is circular to use the natural logarithm if you're trying to prove the formula for the base of that logarithm.

 

[tahayassen]

I expanded that polynomial out. Ignoring the bit about Pascal's triangle, it's straightforward to see that the polynomial expands to:
s_n = n^{-n} (1 + (n-1)n + ... + (n-1)n^{n-1} + n^n)
If you don't believe it, try it for a few small values of n to see it. Now, if you do accept that, then what I said afterwards follows. That is, to a first approximation, we can drop everything but the last two terms in the polynomial, which gives e = 2 after we take the limit. If we keep instead the first three terms, we have
s_n \approx n^{-n} \left(\frac{1}{2}(n^2 -3n +2)n^{n-2} + (n-1)n^{n-1} + n^n\right)
Trying the same strategy, in that first term there is only one term with order n^n, so
s_n \approx n^{-n} (\frac{1}{2} n^n + n^n + n^n) = 2.5.
So the idea is that the more terms you keep, the closer you get to the actual value of e. I think this leads to the Taylor expansion for the exponential function directly, but I haven't worked it out.

By the way -- the "argument" using the natural logarithm is not much of an argument, because you're assuming the nature of e to prove the nature of e. It is circular to use the natural logarithm if you're trying to prove the formula for the base of that logarithm.

I haven't read your post in 100% depth yet, because I just woke up and I need to go to school, but quick question: is there a particular reason for using Sn notation for the left side?

 

[DryRun]

is there a particular reason for using Sn notation for the left side?

S_n is the sequence of partial sums.

Edit: No, it's not. It's just the expansion, using ? theorem.

 

[Steely Dan]

S_n is the sequence of partial sums.

Edit: No, it's not. It's just the expansion, using ? theorem.

I used the s_n notation because of that connection, but it was a weak one because this isn't actually a sum problem, as sharks observes. The point is, whatever is on the right hand side is the current approximation to e.

 

[Mindscrape]

There's a bunch of math that just has loads of tabulated values. e, for example, can be generated by plugging in high enough values of n. sin and cos along with tons of other functions need look up tables, or fast computing algorithms. Younger generations, such as mine, are so used to calculators pumping out values of pi, sin, cos, e, that we forget the nature of these numbers and functions.

 

[tahayassen]

I understand your post Steely Dan, but I was just wondering why I'm not getting the approximate value of 2 when I tried it myself:

http://i2.lulzimg.com/7009a29b9a.png

Tex version:

Ignoring\quad coefficients...\\ { (1+{ n }^{ -1 }) }^{ n }={ \underset { n\rightarrow \infty }{ lim } }(1^{ n }{ ({ n }^{ -1 }) }^{ 0 }+{ 1 }^{ n-1 }{ ({ n }^{ -1 }) }^{ 1 }+{ 1 }^{ n-2 }{ ({ n }^{ -1 }) }^{ 2 }+\quad ...\quad +{ 1 }^{ 1 }{ ({ n }^{ -1 }) }^{ n-1 }+{ 1 }^{ 0 }{ ({ n }^{ -1 }) }^{ n })\\ { (1+{ n }^{ -1 }) }^{ n }={ \underset { n\rightarrow \infty }{ lim } }(1+{ n }^{ -1 }+{ n }^{ -2 }+\quad ...\quad +{ n }^{ -n+1 }+{ n }^{ -n })\\ { (1+{ n }^{ -1 }) }^{ n }={ \underset { n\rightarrow \infty }{ lim } }\left[ { n }^{ -n }({ n }^{ n }+{ n }^{ n-1 }+{ n }^{ n-2 }+\quad ...\quad +{ n }^{ 1 }+{ n }^{ 0 }) \right] \\ Ignoring\quad everything\quad except\quad the\quad first\quad two\quad terms...\\ { (1+{ n }^{ -1 }) }^{ n }={ \underset { n\rightarrow \infty }{ lim } }\left[ { n }^{ -n }({ n }^{ n }+{ n }^{ n-1 }) \right] \\ { (1+{ n }^{ -1 }) }^{ n }={ \infty }^{ -n }(\infty ^{ n }+{ \infty }^{ n-1 })

Results:

http://i2.lulzimg.com/912d8aca39.png

Am I doing something wrong? Do the coefficients matter? How did you get the coefficients of (n-1)?

 

[Steely Dan]

Am I doing something wrong? Do the coefficients matter? How did you get the coefficients of (n-1)?

The coefficients absolutely matter. In the way you wrote it, the second term is not (1)n^{-1} but (n)n^{-1}. If you're not seeing it, you have to actually work out the individual terms in the expansion.

 

[Mark44]

Regarding post #31, you should never substitute ∞ into an arithmetic expression.

In that post you have "∞^-n(∞ⁿ + ∞^-n)"

 

[tahayassen]

s_n = n^{-n} (1 + (n-1)n + ... + (n-1)n^{n-1} + n^n)

How did you get the coefficients: (n-1) for all the terms in the middle?

Wouldn't pascal's coefficients be something like:

http://i2.lulzimg.com/4084d10b33.png

 

[Steely Dan]

s_n = n^{-n} (1 + (n-1)n + ... + (n-1)n^{n-1} + n^n)

How did you get the coefficients: (n-1) for all the terms in the middle?

Wouldn't pascal's coefficients be something like:

http://i2.lulzimg.com/4084d10b33.png

That's correct. Obviously I didn't work out all of the coefficients, just the last three:

\left(\begin{array}{c} n \\ n \end{array}\right) = \frac{n!}{n!\ 0!} = 1
\left(\begin{array}{c} n \\ n-1 \end{array}\right) = \frac{n!}{(n-1)!\ 1!} = n
\left(\begin{array}{c} n \\ n-2 \end{array}\right) = \frac{n!}{(n-2)!\ 2!} = \frac{1}{2}(n)(n-1)

The coefficients are symmetric, so this would also be the first three.

 

[tahayassen]

Thanks! I understand now.

 

[tahayassen]

Actually, one last thing, if I look at your third term:

\frac { 1 }{ 2 } (n^{ 2 }-3n+2)n^{ n-2 }

But shouldn't it be:

\frac { 1 }{ 2 } (n)(n-1)n^{ n-2 }\\ =\frac { 1 }{ 2 } ({ n }^{ 2 }-n)n^{ n-2 }

 

[Steely Dan]

Yes, I tried doing the coefficients in my head when I wrote that and evidently made a mistake. Fortunately it didn't affect the leading term.