How Can Recurrence Relations Help Solve the Hard Logarithm Problem?

[courtrigrad]

Hello all

I need help with the problem attached below. I tried proof by induction, but cannot prove it. P(n) is a polynomial of degree 2n-2. I have to establish the recurrence relation.

Any help is greatly appreciated!

Thanks

 

[mathwonk]

probably you are trying to prove too much, if all you want is to show that all derivatives of e^(-x^(-2)) vanish at zero, but anyway, notice that every time you take a derivative by the product rule, on one hand you differentiate the original function which multiplies that factor by x^(-3), so the degree of that factor goes down by 3. But when you differentiatiate the other factor, the one with negative powers of x, the pwoer of x goes down by 1.

so when you clear denominators, you have to multiply by the larger negative power, which is x^(-3n) where n is the numbwer of tiomes you did this. On the other factor, when you multiply by this, since those powers only went down on e each time,l you get a polynomial in the top whiose degree goes up 2 each time, and it seems form looking at thes econd erivative that it was degree 2 that time, hence ingeneral is degree 2n-2.

i.e. this is obvious from looking at the first two derivatives. I could be wrong of course.

 

[M98Ranger]

for reaching out for help with this problem! The hard problem concerning logarithms is a common one that many students struggle with. Let's break it down step by step to see if we can find a solution together.

First, let's review what we know about logarithms. A logarithm is the inverse function of an exponential function. In other words, if we have an equation in the form of y = a^x, the logarithm is the exponent that we need to raise a to in order to get y. For example, if we have the equation 8 = 2^x, then the logarithm of 8 with base 2 is 3, because 2^3 = 8.

Now, let's take a look at the given problem. We have a polynomial P(n) of degree 2n-2. In order to establish a recurrence relation, we need to find a pattern in the coefficients of the polynomial. Since the degree is 2n-2, we can assume that the polynomial has the form P(n) = a(2n-2) + b(2n-3) + c(2n-4) + ... + k, where a, b, c, ..., k are constants.

Next, we can use the definition of logarithms to rewrite this polynomial as P(n) = a(2^n)^2 + b(2^n)^3 + c(2^n)^4 + ... + k. Now, we can see that the coefficients of the polynomial are increasing by a factor of 2^n each time. This means that we can express the polynomial as P(n) = a(2^n)^2 + b(2^n)^3 + c(2^n)^4 + ... + k = a(2^n)^2 + 2^n(b(2^n)^2) + 2^n(c(2^n)^3) + ... + 2^n(k(2^n)^n-1).

We can simplify this further by factoring out 2^n, giving us P(n) = 2^n(a(2^n) + b(2^n)^2 + c(2^n)^3 + ... + k(2^n)^n-1). Now, we can see that this polynomial has a recursive structure, where each term is multiplied by 2^n. This means that we can establish a recurrence relation as P(n) = 2^nP(n