Studying Mathematics Thread

[polymath257]

I sort of see it now. It seems it's all about continuing the pattern to infinity, starting from an intuitive sawtooth function.

The basic idea is to make sure the problem points get distributed throughout the interval while keeping the function continuous.

In the sin(3^n x)/2^n , the ide is that we have functions of smaller and smaller amplitude but larger and larger derivatives added up in a way that 'problems mount up' at any point.

There is a much deeper result is that *any* continuous function can be approximated as well as you like by a continuous nowhere differentiable function.

[GrandizerII]

Some trigonometry fun:

tanx = sinx/cosx

cotx = cosx/sinx

Is tanx = 1/cotx? And is cotx = 1/tanx?

It seems like the answer to both is "no", but I could be mistaken. Correct me if so.

Here's my [probably mathematically naive] reasoning:

If x = 0, then tanx = sinx/cosx = 0, while 1/cotx = 1/(cosx/sinx) is undefined (since sin(0) = 0). Therefore, tanx != 1/cotx.

If x = pi/2, then cotx = cosx/sinx = 0, while 1/tanx = 1/(sinx/cosx) is undefined (since cos(pi/2) = 0). Therefore, cotx != 1/tanx.

[polymath257]

tan x=1/cot x and cot x=1/tan x *whenever both are defined*.

But the full range of definition of tan x is not the same as the full range of definition of 1/cot x.

[GrandizerII]

The limit (as n approaches infinity) of (1+2+3+...+n)/(n^2) is actually 1/2 ... not 0!

I was confused when I first saw this because just looking at it, it seemed that you just had to increase the value of n to infinity (in which case the limit would be 0), but what I initially failed to keep in mind is that not only is n increasing in value as it approaches infinity but the number of terms increases in the numerator as well. That's why the denominator never goes high enough to demolish the fraction to 0/infinity.

As to how to arrive at the correct answer:

the numerator (1+2+3+...+n) = n(n+1)/2
I will demonstrate this is true another time ...

This means the whole expression up above (after the limit) is n(n+1)/(2n^2) = (n+1)/(2n) = n/(2n)+1/(2n) = 1/2 + 1/(2n)

Therefore, the limit (as n approaches infinity) of that result is 1/2.

[polymath257]

The limit (as n approaches infinity) of (1+2+3+...+n)/(n^2) is actually 1/2 ... not 0!

I was confused when I first saw this because just looking at it, it seemed that you just had to increase the value of n to infinity (in which case the limit would be 0), but what I initially failed to keep in mind is that not only is n increasing in value as it approaches infinity but the number of terms increases in the numerator as well. That's why the denominator never goes high enough to demolish the fraction to 0/infinity.

As to how to arrive at the correct answer:

the numerator (1+2+3+...+n) = n(n+1)/2
I will demonstrate this is true another time ...

This means the whole expression up above (after the limit) is n(n+1)/(2n^2) = (n+1)/(2n) = n/(2n)+1/(2n) = 1/2 + 1/(2n)

Therefore, the limit (as n approaches infinity) of that result is 1/2.

OK, now try

(1^2 + 2^2 + 3^2 +...+n^2)/n^3

and

(1^3 + 2^3 + 3^3 +...+n^3 )/n^4.

Any guesses for a generalization?

[robvalue]

The limit (as n approaches infinity) of (1+2+3+...+n)/(n^2) is actually 1/2 ... not 0!

I was confused when I first saw this because just looking at it, it seemed that you just had to increase the value of n to infinity (in which case the limit would be 0), but what I initially failed to keep in mind is that not only is n increasing in value as it approaches infinity but the number of terms increases in the numerator as well. That's why the denominator never goes high enough to demolish the fraction to 0/infinity.

As to how to arrive at the correct answer:

the numerator (1+2+3+...+n) = n(n+1)/2
I will demonstrate this is true another time ...

This means the whole expression up above (after the limit) is n(n+1)/(2n^2) = (n+1)/(2n) = n/(2n)+1/(2n) = 1/2 + 1/(2n)

Therefore, the limit (as n approaches infinity) of that result is 1/2.

OK, now try

(1^2 + 2^2 + 3^2 +...+n^2)/n^3

and

(1^3 + 2^3 + 3^3 +...+n^3 )/n^4.

Any guesses for a generalization?

After some scrawling and trying to remember old methods, I got

Sum(n^2) = (n/6)(2n^2 + 3n + 1)

Giving limit when divided by n^3 as 1/3.

If I have the energy I'll do n^4 later, but I guess it might be 1/4!

[polymath257]

OK, now try

(1^2 + 2^2 + 3^2 +...+n^2)/n^3

and

(1^3 + 2^3 + 3^3 +...+n^3 )/n^4.

Any guesses for a generalization?

After some scrawling and trying to remember old methods, I got

Sum(n^2) = (n/6)(2n^2 + 3n + 1)

Giving limit when divided by n^3 as 1/3.

If I have the energy I'll do n^4 later, but I guess it might be 1/4!

Good guess (assuming that isn't 4 factorial)! Now the challenge: prove it.

Hint: Riemann

[GrandizerII]

The limit (as n approaches infinity) of (1+2+3+...+n)/(n^2) is actually 1/2 ... not 0!

I was confused when I first saw this because just looking at it, it seemed that you just had to increase the value of n to infinity (in which case the limit would be 0), but what I initially failed to keep in mind is that not only is n increasing in value as it approaches infinity but the number of terms increases in the numerator as well. That's why the denominator never goes high enough to demolish the fraction to 0/infinity.

As to how to arrive at the correct answer:

the numerator (1+2+3+...+n) = n(n+1)/2
I will demonstrate this is true another time ...

This means the whole expression up above (after the limit) is n(n+1)/(2n^2) = (n+1)/(2n) = n/(2n)+1/(2n) = 1/2 + 1/(2n)

Therefore, the limit (as n approaches infinity) of that result is 1/2.

OK, now try

(1^2 + 2^2 + 3^2 +...+n^2)/n^3

and

(1^3 + 2^3 + 3^3 +...+n^3 )/n^4.

Any guesses for a generalization?

First series:

This is the theorem for the numerator series:

1^2 + 2^2 + 3^2 +...+n^2 = n(n+1)(2n+1)/6

Including the denominator, this means the first series up above in the quote is equal to:

n(n+1)(2n+1)/(6n^3) = (n+1)(2n+3)/(6n^2) = (2n^2+5n+3)/(6n^2) = 1/3 + 5/(6n) + 1/(2n^2)

The limit (as n approaches infinity) of n(n+1)(2n+1)/(6n^3) is therefore:

1/3

Second series:

1^3 + 2^3 + 3^3 +...+n^3 = (n^2(n+1)^2)/4

Including the denominator, this means the second series up above in the quote is equal to:

(n^2(n+1)^2)/(4n^4) = (n^2+2n+1)/(4n^2) = 1/4 + 1/(2n) + 1/(4n^2)

The limit (as n approaches infinity) of (n^2(n+1)^2)/(4n^4) is therefore:

1/4

[GrandizerII]

To prove 1+2+3+...+n=n(n+1)/2

Let S = 1+2+3+...+n

In this case, S is also:
n+n-1+n-2+...+1

If you add both equations of S, you get:

2S=n+1+(n-1+2)+(n-2+3) + ... + n+1

2S = n+1 + n+1 + n+1 + ... + n+1 (n times)

Therefore S = n(n+1)/2

Therefore 1+2+3+...+n = n(n+1)/2

Now for the other two theorems stated in my previous post, how to derive them remains a puzzle to me at this stage. I can prove them via mathematical induction only.

[robvalue]

To prove 1+2+3+...+n=n(n+1)/2

Let S = 1+2+3+...+n

In this case, S is also:
n+n-1+n-2+...+1

If you add both equations of S, you get:

2S=n+1+(n-1+2)+(n-2+3) + ... + n+1

2S = n+1 + n+1 + n+1 + ... + n+1   (n times)

Therefore S = n(n+1)/2

Therefore 1+2+3+...+n = n(n+1)/2

Now for the other two theorems stated in my previous post, how to derive them remains a puzzle to me at this stage. I can prove them via mathematical induction only.

I can show you how if you want