Prove that ## \sum_{d\mid n} f(d)=\sum_{d\mid n} f(\frac{n}{d}) ##.

• Math100
In summary: The reason I am taking all this time to explain this is that it seems to me you are trying to jump to the end of the proof without knowing what you are doing.
Math100
Homework Statement
Prove that if ## f ## is any arithmetical function, then ## \sum_{d\mid n}f(d)=\sum_{d\mid n}f(\frac{n}{d}) ##.
[Hint: if you have difficulty with this, write out both sides in the case ## n=10 ##.]
Relevant Equations
None.
Proof:

Let ## n=10 ##.
Observe that
\begin{align*}
&\sum_{d\mid n}f(d)=\sum_{d\mid n}f(\frac{n}{d})\\
&\sum_{d\mid 10}f(d)=\sum_{d\mid 10}f(\frac{10}{d})\\
&f(1)+f(2)+f(5)+f(10)=f(\frac{10}{1})+f(\frac{10}{2})+f(\frac{10}{5})+f(\frac{10}{10})\\
&f(1)+f(2)+f(5)+f(10)=f(10)+f(5)+f(2)+f(1).\\
\end{align*}
Therefore, ## \sum_{d\mid n}f(d)=\sum_{d\mid n}f(\frac{n}{d}) ##.

Math100 said:
Homework Statement:: Prove that if ## f ## is any arithmetical function, then ## \sum_{d\mid n}f(d)=\sum_{d\mid n}f(\frac{n}{d}) ##.
[Hint: if you have difficulty with this, write out both sides in the case ## n=10 ##.]
Relevant Equations:: None.

Proof:

Let ## n=10 ##.
Observe that
\begin{align*}
&\sum_{d\mid n}f(d)=\sum_{d\mid n}f(\frac{n}{d})\\
&\sum_{d\mid 10}f(d)=\sum_{d\mid 10}f(\frac{10}{d})\\
&f(1)+f(2)+f(5)+f(10)=f(\frac{10}{1})+f(\frac{10}{2})+f(\frac{10}{5})+f(\frac{10}{10})\\
&f(1)+f(2)+f(5)+f(10)=f(10)+f(5)+f(2)+f(1).\\
\end{align*}
Therefore, ## \sum_{d\mid n}f(d)=\sum_{d\mid n}f(\frac{n}{d}) ##.
This is the case ##n=10.## What about the general case?

You basically have to show that ##\{d\, : \,d|n\}=\{n/d\, : \,d|n\}.##

fresh_42 said:
This is the case ##n=10.## What about the general case?

You basically have to show that ##\{d\, : \,d|n\}=\{n/d\, : \,d|n\}.##
This is where I don't know how to do, because I got the case ## n=10 ## from the hint. But how do I show that those two are equal by proving them with the general case?

A common way to show the equality of sets is to show that they are contained in each other.

a) Say ##d|n.## Then ##n=d\cdot e## for some ##e## and ##e=\dfrac{n}{d} \,|\,n.## This shows ##\{n/d\, : \,d|n\}\subseteq \{d\, : \,d|n\}## since we showed that ##e## is a divisor of ##n##.

b) Say ##e=\dfrac{n}{d}## for some divisor ##d## of ##n.## Then ##e\cdot d= n## and therefore ##e|n.##
This shows ##\{n/d\, : \,d|n\}\subseteq \{d\, : \,d|n\}.##

From a) and b) we get ##\{n/d\, : \,d|n\} = \{d\, : \,d|n\}.##

It is a bit tricky to see that this is actually a proof because the sets are so similar. You could try to prove the following with that method:

A number ##n## is said to be odd if it can be written as ##n=2k+1.##
A number ##n## is said to be even if it is not odd.
Prove that even numbers are exactly those that can be divided by ##2,## i.e. show that
$$\{n\, : \,n \text{ is even }\}=\{n\, : \, 2\,|\,n\}$$
You have to use the definitions here.

fresh_42 said:
A common way to show the equality of sets is to show that they are contained in each other.

a) Say ##d|n.## Then ##n=d\cdot e## for some ##e## and ##e=\dfrac{n}{d} \,|\,n.## This shows ##\{n/d\, : \,d|n\}\subseteq \{d\, : \,d|n\}## since we showed that ##e## is a divisor of ##n##.

b) Say ##e=\dfrac{n}{d}## for some divisor ##d## of ##n.## Then ##e\cdot d= n## and therefore ##e|n.##
This shows ##\{n/d\, : \,d|n\}\subseteq \{d\, : \,d|n\}.##

From a) and b) we get ##\{n/d\, : \,d|n\} = \{d\, : \,d|n\}.##

It is a bit tricky to see that this is actually a proof because the sets are so similar. You could try to prove the following with that method:

A number ##n## is said to be odd if it can be written as ##n=2k+1.##
A number ##n## is said to be even if it is not odd.
Prove that even numbers are exactly those that can be divided by ##2,## i.e. show that
$$\{n\, : \,n \text{ is even }\}=\{n\, : \, 2\,|\,n\}$$
You have to use the definitions here.
Suppose ## n ## is even.
Then ## n=2a ## for some ## a\in\mathbb{Z} ##.
This implies ## 2\mid n ##.
Thus ## \{n\, : \,n \text{ is even }\}=\{n\, : \, 2\,|\,n\} ##.

Math100 said:
Suppose ## n ## is even.
Then ## n=2a ## for some ## a\in\mathbb{Z} ##.
Yes, but why? More detailed:

##n## is even, i.e. not odd. So ##n\neq 2k+1##. Since ##2(k-1)+1=2k-1## and ##2(k+1)+1=2k+3## are all odd per definition, we can only have ##2(k-1)+0,2k+0,2(k+1)+0## within the gaps of odd numbers for ##n.## Thus ##n=2k## for some ##k## and so ##2\,|\,n.##
Math100 said:
This implies ## 2\mid n ##.
Thus ## \{n\, : \,n \text{ is even }\}=\{n\, : \, 2\,|\,n\} ##.
The other direction? So far, we have shown ## \{n\, : \,n \text{ is even }\}\subseteq \{n\, : \, 2\,|\,n\}. ##

Now assume ##2\,|\,n.## Then ##n=2\cdot e## for some ##e.## If ##n## was odd, then ##2e=2k+1## for some ##k.## Therefore ##2(e-k)=1## and ##2\,|\,1## which is impossible. Hence, ##n## cannot be odd, which is the definition of even, i.e. ## \{n\, : \,n \text{ is even }\}\supseteq \{n\, : \, 2\,|\,n\}. ##

Both inclusions ## \{n\, : \,n \text{ is even }\}\subseteq \{n\, : \, 2\,|\,n\} \subseteq \{n\, : \,n \text{ is even }\} ## imply equality.

You cannot just jump to what you want to show. Read the definition! I defined 'even' as 'not odd', and 'odd' as ##2k+1.## I did not define 'even' as ##2k.## This had to be proven.

This is an almost trivial example, admitted, but it should show you the steps from what we know to what we want to prove.

Math100

1. What is the purpose of this equation?

This equation is used to prove the arithmetic function identity known as the Dirichlet convolution. It is commonly used in number theory and has applications in various mathematical fields.

2. How do you interpret the notation used in this equation?

The sigma symbol (Σ) represents the summation of the values of the function for all divisors of n. The "d|n" notation means that d is a divisor of n. The function f(d) is evaluated for each divisor d of n and the results are summed together.

3. Can you provide an example of how this equation is used?

Sure, let's say we have the function f(d) = d^2 and we want to prove that the equation holds for n = 6. The divisors of 6 are 1, 2, 3, and 6. So, the left side of the equation would be f(1) + f(2) + f(3) + f(6) = 1^2 + 2^2 + 3^2 + 6^2 = 50. The right side of the equation would be f(6/1) + f(6/2) + f(6/3) + f(6/6) = f(6) + f(3) + f(2) + f(1) = 6^2 + 3^2 + 2^2 + 1^2 = 50. As you can see, both sides of the equation equal 50, thus proving the identity for n=6.

4. What are some other applications of this equation?

This equation has various applications in number theory, such as finding the number of divisors of a given number, calculating the Euler totient function, and solving problems related to multiplicative functions. It also has applications in algebraic geometry, representation theory, and combinatorics.

5. Are there any special cases where this equation does not hold?

Yes, there are some special cases where this equation may not hold. For example, if the function f(d) is not multiplicative, meaning that f(ab) is not equal to f(a)f(b) for all integers a and b, then the equation may not hold. Also, if the function is not defined for certain values, the equation may not hold. It is important to check the conditions and assumptions for the specific use of this equation.

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