Find the derivative of the following function:
To take this derivative we want to use logarithmic differentiation. To that end we take the derivative of both sides,
Therefore, taking the derivative of both sides, we have
Find the derivative of the following function:
To take this derivative we want to use logarithmic differentiation. To that end we take the derivative of both sides,
Therefore, taking the derivative of both sides, we have
If for , what conditions are needed for the inequality
to hold?
by the inductive hypothesis. But then, since ,
Thus, the statement is true for ; and hence, for all
Prove the identity:
for .
While, on the right we have,
So, the identity holds in the case . Assume then that it holds for some . Then we have,
Hence, the statement is true for , and so, for all
Use induction to prove that if for all , then
This is the telescoping property for products.
Thus, the property holds for the case . Assume then that it holds for some . Then,
Hence the property is true for ; and thus, for all
Prove the multiplicative property of the product, i.e.,
Thus, the multiplicative property holds for the case . Assume then that it holds for some . Then,
Thus, the case is true; therefore, the property holds for all
Give an inductive definition of the symbol
Claim:
Proof. If , we have on the left, and on the right we have . Thus, the formula is true for .
Assume then that the formula is true for some . Then,
Thus, if the formula is true for then it is true for . Since we have established that it is true for , we then have that it is true for all
Update: From a request in the comments, we’ll add in a way to arrive at the formula (without just guessing).
First, we write,
Then, we consider the product
Where in the last line we cancelled terms again. The only things we are left with are the in the numerator and the 2 and in the denominator. Of course, this is pretty much a proof that the formula is correct without using induction, but it doesn’t rely on us guessing the formula correctly.
As noted in the comments, often it is easier to guess the correct formula and use induction to prove the formula is correct than to derive the formula directly.
Claim:
Proof. If on the left we have and on the right we have . Thus, the formula is true for the case .
Assume then that the formula is true for some . So,
Thus, if the formula is true for then it is true for . Since we have established that it is true for , we have that is true for all