Question

Calculate enough derivatives to guess the general formula for `f^((n))(x)`. And proof this by mathematical induction. `f(x) = 1/2-x` and `f(x) = cos(ax)`?

For the first one I come to a formula `n!(x-x)^-(n+1)`

This is okay, but I dont understand de proof in the book. The following solution is written.

Assume `f^(k) (x) = k!(2 − x)^(−(k+1))` (i.e., (*) holds for n = k)

Then `f^(k+1)(x) = k!(−(k +1)(2− x)^(−(k+1)−1)(−1))` What is this ???

= `(k +1)!(2−x)^(−((k+1)+1))`. This is the formule like I had but then instead of n, (k+1). I understand that, but not the other one.

For the second question I calculated till the 5th deriavative. But I dont see a general formule because the sin and cos are changing every time.

Answer 1

`(A) \ \ (d^n)/(dx^n) 1/(2-x) = n!(2-x)^(-(n+1)) = (n!)/(2-x)^(n+1)`

`(B) \ \ (d^n)/(dx^n) cos(ax) = a^ncos(ax+(npi)/2)`

Explanation:

We seek a proven result for the `n^(th)` derivative for:

`(A) \ \ f(x)=1/(2-x)`
`(B) \ \ f(x)=cos(ax)`

Part (A):

`f(x)=1/(2-x)`

Construction of the Relationship

Writing the function as `f(x)=(2-x)^(-1)` and applying the power rule we can form the first few derivative:

`f^((1))(x) = (-1)(2-x)^(-2)(-1) = (2-x)^(-2)` ..... [A1]
`f^((2))(x) = (-2)(2-x)^(-3)(-1) = 2(2-x)^(-3)`
`f^((3))(x) = 2(-3)(2-x)^(-4)(-1) = 2.3(2-x)^(-4)`
`f^((3))(x) = 2.3(-4)(2-x)^(-4)(-1) = 2.3.4(2-x)^(-5)`

Leading us to propose the `n^(th)` derivative is:

`f^((n))(x) = n!(2-x)^(-(n+1)) = (n!)/(2-x)^(n+1)`

We can test this hypothesis using Mathematical Induction, as follows:

Induction Proof - Hypothesis

We seek to prove that:

`f^((n))(x) = (d^n)/(dx^n) 1/(2-x) = n!(2-x)^(-(n+1))` ..... [B1]

So let us test this assertion, [B1], using Mathematical Induction:

Induction Proof - Base case:

We have already shown that the given result, [B1], holds for `n=1` using [A1]

So the given result is true when `n=1`.

Induction Proof - General Case

Now, Let us assume that the given result [B1] is true when `n=m`, for some `m in NN, m gt 1`, in which case for this particular value of `m` we have:

`f^((m))(x) = (d^m)/(dx^m) 1/(2-x) = n!(2-x)^(-(n+1))` ..... [C1]

Then, differentiating [C1] wrt `x`, by applying the power rule, we have:

`f^((m+1))(x) = (d^(m+1))/(dx^(m+1)) 1/(2-x)`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (-(n+1))n!(2-x)^(-(n+1)-1)(-1)`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (n+1)n!(2-x)^(-(n+1+1))`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (n+1)!(2-x)^(-((n+1)+1))`

Which is the given result [B1] with `n=m+1`

Induction Proof - Summary

So, we have shown that if the given result [B1] is true for `n=m`, then it is also true for `n=m+1`. But we initially showed that the given result was true for `n=1` so it must also be true for `n=2, n=3, n=4, ...` and so on.

Induction Proof - Conclusion

Then, by the process of mathematical induction the given result [B1] is true for `n in NN`

Hence we have:

`f^((n))(x) = (d^n)/(dx^n) 1/(2-x) = n!(2-x)^(-(n+1)) \ \ \` QED

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Part (B):

`f(x)=cos(ax)`

Construction of the Relationship

We can form the first few derivative:

`f^((1))(x) = -asin(ax)` ..... [A2]
`f^((2))(x) = a^2cos(ax)`
`f^((3))(x) = -a^3sin(ax)`
`f^((3))(x) = a^4cos(ax)`

Using the cosine sum of angle formula:

`cos (A+B) -= cosAcosB - sinAsinB`

And, exploiting the phase shift properties, we have:

`cos(x+(npi)/2) = cosx cos((npi)/2) - sinx sin((npi)/2)`

Leading us to propose the `n^(th)` derivative is:

`f^((n))(x) = a^ncos(ax+(npi)/2)`

We can test this hypothesis using Mathematical Induction, as follows:

Induction Proof - Hypothesis

We seek to prove that:

`f^((n))(x) = (d^n)/(dx^n) cos(ax) = a^ncos(ax+(npi)/2)` ..... [B2]

So let us test this assertion, [B2], using Mathematical Induction:

Induction Proof - Base case:

We have already shown that the given result, [B2], holds for `n=1` using [A2]

So the given result is true when `n=1`.

Induction Proof - General Case

Now, Let us assume that the given result [B2] is true when `n=m`, for some `m in NN, m gt 1`, in which case for this particular value of `m` we have:

`f^((m))(x) = (d^m)/(dx^m) cos(ax) = a^mcos(ax+(mpi)/2)` ..... [C2]

Then, differentiating [C2] wrt `x`, we have:

`f^((m+1))(x) = (d^(m+1))/(dx^(m+1)) a^mcos(ax+(mpi)/2)`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = a^m(a)(-1)sin(ax+(mpi)/2)`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = -a^(m+1)sin(ax+(mpi)/2)`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = a^(m+1)cos(ax+((m+1)pi)/2)`

Which is the given result [B2] with `n=m+1`

Induction Proof - Summary

So, we have shown that if the given result [B2] is true for `n=m`, then it is also true for `n=m+1`. But we initially showed that the given result was true for `n=1` so it must also be true for `n=2, n=3, n=4, ...` and so on.

Induction Proof - Conclusion

Then, by the process of mathematical induction the given result [B2] is true for `n in NN`

Hence we have:

`f^((n))(x) = (d^n)/(dx^n) cos(ax) = a^ncos(ax+(npi)/2) \ \ \` QED