Showing that e^a * e^b = e^(a+b)

  • Thread starter dacruick
  • Start date
In summary: I apologize for any confusion or inconvenience that this may have caused.In summary, the author attempted to solve a problem where two infinite series must be equal, but does not know how to do so without expanding the series. They have attempted to multiply each series by the last two terms, but are unsure if this is sufficient.
  • #1
dacruick
1,042
1

Homework Statement


By definition, [itex]e^{z}[/itex] = [itex]\sum[/itex][itex]\frac{1}{n!}[/itex][itex]z^{n}[/itex]

Use this to show the relationship in the question title


The Attempt at a Solution


Well, what I've tried to do is as follows:
[itex]e^{z_{1}}[/itex][itex]e^{z_{2}}[/itex] = [itex]\sum[/itex][itex](z_{1}z_{2})^{n}[/itex]/[itex]{n!^{2}}[/itex]

And set that equal to
[itex]e^{(z_{1}+z_{2})}[/itex] = [itex]\sum[/itex][itex]\frac{1}{n!}[/itex][itex](z_{1}+z_{2})^{n}[/itex]

What I'm left with is this expression that
[itex](z_{1}+z_{2})^{n}[/itex] = [itex](z_{1}z_{2})^{n}/n![/itex]

those are all sums from n=0 to n=∞ still btw. Sorry for my LameTex inexperience.

My question is, is there any way that I can show that those two are equal? Or have I gone about this all wrong. Thanks!

dacruick
 
Physics news on Phys.org
  • #2
You can't multiply infinite series like you've done. In fact you can't even multiply finite polynomials like that. For example consider the multiplication of the following quadratics:

[tex](x^2 + x + 1)(y^2 + y + 1) = (xy)^2 + x^2y + xy^2 + x^2 + y^2 + x + y + 1 \neq (xy)^2 + xy + 1[/tex]
 
  • #3
Poopsilon said:
You can't multiply infinite series like you've done. In fact you can't even multiply finite polynomials like that. For example consider the multiplication of the following quadratics:

[tex](x^2 + x + 1)(y^2 + y + 1) = (xy)^2 + x^2y + xy^2 + x^2 + y^2 + x + y + 1 \neq (xy)^2 + xy + 1[/tex]

I'm so stupid...Haha, not just because of the foolishness you stated but because as soon as you said that I'm pretty sure I figured out how to do it. Maybe it had something to do with that cup of coffee I just made. You'll have to tussle with Starbucks for partial credit of my pending success.
 
  • #4
Starbucks is far too successful already, I want all of it.
 
  • #5
Haha get your lawyer ready then :P.

So I have the answer, I've expanded both of the series for n = 0, 1, 2, and 3 and I can sufficiently see that the terms will cancel out on both sides.

However, I question the completeness of my answer. I assume that the only way to get marks for this badboy is to expand both of the series' in terms of 'n'. I don't readily see how to do this. Would it work if I just multiplied each of the first 4 values of 'n' by the last 2 terms of each series, and then see if those cancelled?

I have next to no pure mathematical background, and I'm unclear as to what constitutes "showing" that these two series are equal.
 

1. How do you prove that e^a * e^b = e^(a+b)?

The proof involves using the properties of exponents and the definition of the natural logarithm. We can rewrite e^a as e^(a*ln(e)) and e^b as e^(b*ln(e)). Using the property (e^x)^y = e^(xy), we can simplify e^(a*ln(e)) * e^(b*ln(e)) to e^(a*ln(e) + b*ln(e)). Finally, using the property ln(xy) = ln(x) + ln(y), we can simplify e^(a*ln(e) + b*ln(e)) to e^(ln(e^(a+b))). Since ln(e^(a+b)) = a + b, we get the desired result e^(a+b).

2. What is the significance of the number e in this equation?

The number e is a mathematical constant that is approximately equal to 2.71828. It is the base of the natural logarithm and appears frequently in exponential functions. In this equation, e represents the base of the exponential function and is used to simplify the expression on the left side to the expression on the right side.

3. Can this equation be used for any values of a and b?

Yes, this equation is valid for any real values of a and b. This can be seen by plugging in different values for a and b and verifying that the equation holds true.

4. Why is the equation written as e^a * e^b = e^(a+b) and not e^(a+b) = e^a * e^b?

This is due to the commutative property of multiplication. According to this property, the order of multiplication does not affect the result. Therefore, both expressions are equivalent and can be written in either form.

5. How is this equation used in real-life applications?

This equation is commonly used in fields such as physics, engineering, and finance to model exponential growth or decay. It is also used in probability and statistics to calculate the probability of independent events occurring simultaneously. Additionally, it is used in population dynamics and other biological processes to model the growth of populations over time.

Similar threads

Derivative of Determinant of Metric Tensor With Respect to Entries
    • Calculus and Beyond Homework Help
Replies
14
Views
3K
I On a remark regarding the Cauchy integral formula
    • Topology and Analysis
Replies
2
Views
635
Center of mass of spherical shell inside of cone (Apostol Problem).
    • Calculus and Beyond Homework Help
Replies
3
Views
556
A Papapetrou transformation: Conditions to be satisfied to achieve transformation
    • Special and General Relativity
Replies
1
Views
789
Understanding the Cauchy Integration Formula for Analytic Functions
    • Calculus and Beyond Homework Help
Replies
4
Views
1K
B Calculating the Total Energy of a lattice w/ the Madelung constant
    • Classical Physics
Replies
1
Views
1K
Finding a periodic solution to ODE
    • Calculus and Beyond Homework Help
Replies
16
Views
558
Show that the limit (1+z/n)^n=e^z holds
    • Calculus and Beyond Homework Help
Replies
6
Views
472
On pointwise convergence of Fourier series
    • Calculus and Beyond Homework Help
Replies
3
Views
411
Groups that cannot be the direct product of subgroups
    • Calculus and Beyond Homework Help
Replies
1
Views
1K

Hot Threads

Recent Insights

Back
Top