Proving Exponential Function Properties with Inequalities | Homework Solution

[mynameisfunk]

Homework Statement

For the exponential function $exp(z) = \sum^{\infty}_{n=0} \frac{z^n}{n!}$, we know that $exp(z+w)=exp(z).exp(w)$, and that, if $x \geq 0$, then $exp(x) \geq 1+x$. Use these facts to prove that, if $1 \leq n \leq m$, then

$$|\prod^n_{j=1} (1+a_j) - \prod^m_{j=1} (1+a_j)| \leq exp(\sum^n_{j=1} |a_j|) . (exp(\sum^m_{j=n+1} |a_j|)-1)$$

Homework Equations

The Attempt at a Solution

I simplified to:

$$|a_{n+1}+a_{n+2}+...+a_m+a_{n+1}a_{n+2}+...+a_{n+1}a_{n+2}...a_m| \leq \sum^{\infty}_{n=0} \frac{(\sum^m_{j=1} |a_j|)^n}{n!} - \sum^{\infty}_{n=0} \frac{(\sum^n_{j=1} |a_j|)^n}{n!}$$

from here I had the idea that I could break n=0 and n=1 out of the exponentials since n=0 doesn't yield anything and n=1 yields some individual terms i could subtract from the left side. I am thinking this is the wrong approach though

 

[mighty2000]

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I would first double check the given facts to make sure they are correct. The first fact, exp(z+w)=exp(z).exp(w), is a well-known property of exponential functions and can be easily verified by plugging in values for z and w. The second fact, exp(x)≥1+x for x≥0, is also true and can be proved using the Maclaurin series expansion of exp(x).

Next, I would try to understand the problem by looking at the given equation and seeing what it represents. The left side is the difference between two products, each with n and m factors respectively. The right side involves two exponential functions with sums of absolute values of the a_j terms. This suggests that the problem may involve some kind of approximation or estimation.

To prove the given inequality, I would start by using the first fact to rewrite the left side of the equation as:

|∏^n_{j=1}(1+a_j) - ∏^m_{j=1}(1+a_j)| = |∏^n_{j=1}(1+a_j) - ∏^n_{j=1}(1+a_j)∏^{m-n}_{j=1}(1+a_{n+j})|

Next, I would use the second fact to rewrite the second term on the right side as:

|∏^n_{j=1}(1+a_j)∏^{m-n}_{j=1}(1+a_{n+j})| ≤ exp(∑^n_{j=1}|a_j|)exp(∑^{m-n}_{j=1}|a_{n+j}|)

Now, I would expand the exponentials using their Maclaurin series expansions and group the terms with the same powers of ∑^n_{j=1}|a_j| and ∑^{m-n}_{j=1}|a_{n+j}|. This will give a series of terms on the right side, each with a coefficient that is a product of some combination of n and m terms. For example, the term with ∑^n_{j=1}|a_j|^n will have a coefficient of 1/n!. The remaining terms will have coefficients that are products of n and m terms, such as (n+m-1)/(n!(m-1)!).

Finally, I would use the fact that n