recursive function for Exponential function

[dimakvirik]

I connect exponential function to recursive. I don't thing I invent something new and maybe for this formula already exist some name?

 

[dimakvirik]

If you put a=1 and b =0. on final formula will left only derivative

 

[dimakvirik]

derivative can be replaced to n!

 

[stapel]

View attachment 34860

\(\displaystyle f_{0} (x) = x^n\)
\(\displaystyle f_{n-1} = f_{n-2}(x+a) - f_{n-2}(x+b)\)
\(\displaystyle f_n = x^n \times \frac{d}{dx^n} (a-b)^n\)
\(\displaystyle n > 1\)

I connect exponential function to recursive. I don't thing I invent something new and maybe for this formula already exist some name?

If you put a=1 and b =0. on final formula will left only derivative

derivative can be replaced to n!

Note: \(\displaystyle y=x^n\) is a polynomial function, not an exponential function.

What was the original function, to which you have "connected" this recursive function?

Where do "a" and "b" come from?

How did you go from the difference expression for \(\displaystyle f_{n=1}\) to the product expression for \(\displaystyle f_n\)?

If a=0, b=0, then \(\displaystyle f_n = 0\) for all x. Exponential functions do not have derivatives that are uniformly equal to zero.

Assuming that "a" and "b" are constants, the derivative of their difference will always be zero, regardless of their own values. Again, this is not true of exponential functions (o rof functions in general). Was that what you meant?

Thank you!

Eliz.

 

[Steven G]

\(\displaystyle f_{0} (x) = x^n\)
\(\displaystyle f_{n-1} = f_{n-2}(x+a) - f_{n-2}(x+b)\)
\(\displaystyle f_n = x^n \times \frac{d}{dx^n} (a-b)^n\)
\(\displaystyle n > 1\)

Eliz, thanks for making us a cleaner version!

Re: \(\displaystyle f_n = x^n \times \frac{d}{dx^n} (a-b)^n\)

Since, \(\displaystyle (a-b)^n\) is a constant function of x, we get \(\displaystyle \frac{d}{dx} (a-b)^n=0\).

Then \(\displaystyle \frac{d}{dx^n} (a-b)^n\)=0.

This makes \(\displaystyle f_n = x^n \times \frac{d}{dx^n} (a-b)^n = 0\)

 

[dimakvirik]

This makes fn=xn×ddxn(a−b)n=0\displaystyle f_n = x^n \times \frac{d}{dx^n} (a-b)^n = 0fn=xn×dxnd(a−b)n=0

a and b it is constants right. I think I've written in wrong form. so I will try to explain more detail.
[math]f_{0}=x^{n}[/math] it is a polynomial function
than if we take difference between two side number we will get this formula [math]f_{n-1}=f_{n-2}(x+1)-f_{n-2}(x)[/math] for any n>1
than we get [math]f_{n}=\frac{d}{dx^n}x^n=n!=f_{n-1}(x+1)-f_{n-1}(x)[/math]example:
[math]f_{0}=x^{3}[/math]let's take x=4

[math]f_{1}=f_{0}(x+1)-f_{0}(x)=125-64=61[/math][math]f_{2}=f_{1}(x+1)-f_{1}(x)=91-61=30[/math][math]f_{3}=f_{2}(x+1)-f_{2}(x)=36-30=6=3![/math]
than if we thing what will be if we take difference between a and b we will get that equition which I wrote at a top.
it is right for every n in N set and x in R

 

[topsquark]

a and b it is constants right. I think I've written in wrong form. so I will try to explain more detail.
[math]f_{0}=x^{n}[/math] it is a polynomial function
than if we take difference between two side number we will get this formula [math]f_{n-1}=f_{n-2}(x+1)-f_{n-2}(x)[/math] for any n>1
than we get [math]f_{n}=\frac{d}{dx^n}x^n=n!=f_{n-1}(x+1)-f_{n-1}(x)[/math]example:
[math]f_{0}=x^{3}[/math]let's take x=4

[math]f_{1}=f_{0}(x+1)-f_{0}(x)=125-64=61[/math][math]f_{2}=f_{1}(x+1)-f_{1}(x)=91-61=30[/math][math]f_{3}=f_{2}(x+1)-f_{2}(x)=36-30=6=3![/math]
than if we thing what will be if we take difference between a and b we will get that equition which I wrote at a top.
it is right for every n in N set and x in R

You still haven't finished defining things!

This a and b thing is confusing and you don't need it in the form you just posted above.

And I haven't finished working on it, but it would appear that it only works for x = n + 1.

Please make sure to post the whole question accurately, otherwise we can't help you properly!

-Dan