• C++ Programming for Financial Engineering
    Highly recommended by thousands of MFE students. Covers essential C++ topics with applications to financial engineering. Learn more Join!
    Python for Finance with Intro to Data Science
    Gain practical understanding of Python to read, understand, and write professional Python code for your first day on the job. Learn more Join!
    An Intuition-Based Options Primer for FE
    Ideal for entry level positions interviews and graduate studies, specializing in options trading arbitrage and options valuation models. Learn more Join!

Simple Quiz

kean

Mathematics Student
Joined
5/31/06
Messages
246
Points
28
Is there a number that is exactly 1 more than its cube?

Show your proof if your answer is yes or no.:dance:
 
There is just one real number:

\(f(x) = x^3-x+1=0\)
\(f'(x)=3x^2-1=0\)
\(x_1=\frac{1}{sqrt(3)} x_2=-\frac{1}{sqrt(3)}\)
\(f(\frac{1}{sqrt(3)}) = 0.615100\) - minimum
\(f(-\frac{1}{sqrt(3)}) = 1.384900\) - maximum
\(f(-infinity) = -infinity\) it means you have a solution for f(x)=0

such solution is close to
x = -1.3247179574971
with approximation error = .0000000010761969093664

-V-
 
Not quite right

Sorry mate. Not quite right. Hint: an undergraduate type question based on Intermediate Value Theorem.

There is just one real number:

\(f(x) = x^3-x+1=0\)
\(f'(x)=3x^2-1=0\)
\(x_1=\frac{1}{sqrt(3)} x_2=-\frac{1}{sqrt(3)}\)
\(f(\frac{1}{sqrt(3)}) = 0.615100\) - minimum
\(f(-\frac{1}{sqrt(3)}) = 1.384900\) - maximum
\(f(-infinity) = -infinity\) it means you have a solution for f(x)=0

such solution is close to
x = -1.3247179574971
with approximation error = .0000000010761969093664

-V-
 
FYI

Andy,

The question did not ask to find the root. I put this question here because I did exactly the same like your answer plus Vadim's answer.
My lecturer raised a good question: Do you need to answer what I do not ask for?

The main issue is understanding of the question. So, do we really understand the market when we do maths or modeling??

I leave it to you. What the question is actually ask for is more important in this aspect. I did not ask to find the root.

Cheers,
 
No intention to challenge anyone. This is merely a knowledge sharing post with due respect.
Thanks.
 
Back
Top