Hello everyone,

I'm pretty stuck with a proof assignment of mine, which states that
1+x+x² /(2!)+...+x²ⁿ /(2n)!=0, n> =0 does not have any real roots.

I know the keyword for this is induction,
but I have no idea how to continue with the
induction step after I've done the basic step and
induction assumption (1+x+...+x²ⁿ /(2n)!=0 doesn't have any real roots).

Please help me out,
Marjo

As it's an assignment, I don't think I can tell you from the start much more than to look at the derivative and specifically the double derivative of this function. See what things you can then draw intuitionally, next find a rigorous argument for them...

Write again if you tried my hints and didn't get anywhere.

Marjo,

I shall assume you have done some calculus and know the second derivative represents the convexity of a function. If you don't understand these, write back and I'll explain them.

To ease my typing, I will set P_n (x)=1+x+x² /2+...+x²ⁿ /(2n)!. Now our claim is P_n (x) does not have any real roots for all non-negative integer n.

First, the case n=0 is trivial.

Now suppose P_k-1 (x) is true (k> 0). We note that P_k ''(x)=P_k-1 (x). But we know P_k-1 (x) doesn't have any real root, so P_k-1 (x)> 0 for all x. This means the graph of y=P_k (x) is concave upwards (i.e. convex). Hence all tangents lie below the the curve. Find the (unique) stationary point of P_k (x). The final step is pretty easy to spot and I'll leave it to you.

Kerwin

Interesting - I have seen this before in a magazine (it was, you may be interested to know, a problem in the selection round of the 1995 Estonian Mathematical Olympiad), but I know of no elementary proof by induction. However, it can be proved rather elegantly using Taylor's theorem. Do you know this result?

David

This makes it obvious that the polynomial is always positive.

There is a reasonably elementary (no calculus) proof using the "Upper and Lower Bound Rule" and simply using a recurrance relation to prove that for x > -2n, P_n (x) > P_n-1 (x), which by assumption, is greater than 0. I didn't mention it because I would think the upper and lower bound rule is nonstandard, and, to be honest, I have no idea how to prove it. I don't think it's as elegant as the proof I hinted to above, but it doesn't require calculus. If anyone's interested though, I can show the proof...

Brad

This question is to Kerwin. I follow your argument all the way to when you attempt to find the Unique stationary point for P_k x. I differentiated P_k x, to get
1+x+...+x^2n-1 which I put equal to 0 for the stationary point, how do I know that the value of x given by the solution of this polynomial is a unique value for the stationary point? Even then, I then have to go on to substitute this value of x into the original equation to show that it doesn't go below the x axis - how do I find the value of this function at the x value in question?

Amars,

The answer is - you don't find the value of the root of P_k '(x). You only need to note that this root is strictly negative. This root is unique because P_k ''(x) is never zero and is strictly positive, hence P_k '(x) is strictly increasing. Let z be this root. Now we note that P_k (x)º P_k '(x)+x²ⁿ /(2n)! and the rest is simple.

Kerwin

Thank you all for your great advice!
I believe I can now continue with this.

Thanks,
Marjo