Mind your Decisions, I'm Presh Talwalkar.
π,π,π,π.
Write an equation that uses π exactly 4 times and equals 0.
Do the same thing for 1 and 2, and see how high you can go.
For a real challenge, solve for any integer n.
You can use common mathematical operations
limited to the following table.
These are standard functions you would find on an Android calculator.
I thank many people who suggested similar problems,
including Tai C. Chiang, Peter,
and the Puzzling StackExchange community.
Pause the video if you'd like to give this problem a try, and when you're ready, keep watching to learn how to solve this problem.
We'll get started with an equation for 0.
Here's one of the many equations you could've written.
π + π - π - π=0.
From here, rather than writing formulas for specific numbers,
let's write a general formula for any positive integer n.
To do that, we'll solve a similar problem.
How can you make n from 3 copies of 2?
There's a clever solution from the Nobel laureate physicist Paul Dirac.
Here was his formula.
Let's write the following:
-log base 2 of log base 2.
Now we are going to iterate the square root function a total of n times.
In the end ,we're going to evaluate √2. This exactly equals n. Why is that?
Let's understand the formula by working through this nested square root.
Recall that √2 = 2^(1/2).
Therefore, √(√2) = 2^(1/2)^2.
If we iterate the square root function a total of n times,
we'll end up with 2^(1/(2^n)).
This is also written as 2^(2^-n).
We now evaluate log base 2 of this, which will be equal to 2^-n.
Evaluating the log base 2 again, we get -(-n), and the opposite of -n is n;
therefore, this formula is exactly true.
Now, let's solve our problem with π.
One thing we can do is we can change the base of this logarithm and the
argument in this nested square root to both be π.
Both of these will cancel out.
We can then rewrite 2 as equaling (- cos π) - cos π.
We thus have a valid expression that uses π a total of 4 times.
This will give all positive integers n.
We can change the negative sign into a positive to get all negative values of n.
But we have one problem:
the logarithm function is using a base of 2.
Log base 2 doesn't appear on a standard Android calculator.
So, how do we get around this?
We'll use the change of base formula for logarithms
log base 2 of x = ln x/ln 2.
We'll use this formula ,and substitute n.
We end up with the following fraction.
Now here we have log base π, so we'll again use the change of base formula.
So now let's substitute that in, and put everything in terms of the natural logarithm.
Now, we have a negative sign at the beginning, which we can cleverly eliminate by taking the reciprocal of this argument.
This will make the formula a little bit easier to read.
We now have a formula for all positive integers n.
Then, if we want to get all negative integers, we simply have a negative sign.
You can also see that if n = 0, this entire thing will actually evaluate to be 0 as well.
So we have one simple formula, and it uses π exactly four times.
Incredible!
Thanks for making Mind your Decisions one of the best channels on YouTube. As always, thanks for watching, and thanks for your support.