# Power Series?

1. Aug 7, 2009

### icystrike

1. The problem statement, all variables and given/known data
Given f(x)=a0x0+a1x1+a2x2...anxn
p.s: anxn is to represent a sub n multiply x^n
f(1)=8 ,
f(35)=e^e
Find f(7)

I dont know what are the theorem that i have to know and how can i proceed

2. Relevant equations

3. The attempt at a solution

2. Aug 7, 2009

### Дьявол

Do you mean:

$$f(x)=a_0x^0+a_1x^1+a_2x^2+...+a_nx^n$$

It is actually:

$$\sum_{i=0}^{n}a_ix^i$$

but doesn't matter. The point is:

$$8=a_0+a_1+a_2+...+a_n$$

and

$$e^e=a_035^0+a_135^1+a_235^2+...+a_n35^n$$

$$y=a_07^0+a_17^1+a_27^2+...+a_n7^n$$

Can you continue now?

P.S 35 = 7 * 5

Last edited: Aug 7, 2009
3. Aug 7, 2009

### icystrike

Thanks Дьявол!
How do I proceed from here?

$$e^e=a_05^07^0+a_15^17^1+a_25^27^2+...+a_n5^n7^n$$

Last edited: Aug 7, 2009
4. Aug 7, 2009

### Staff: Mentor

Is your function in fact a finite sum? IOW, is it defined this way (finite sum)?
$$f(x) = a_0 + a_1 x + a_2 x^2 + ... + a_n x^n$$
or this way (infinite sum)?
$$f(x) = a_0 + a_1 x + a_2 x^2 + ... + a_n x^n + ...$$

The reason I ask is that there is an infinite sum representation for ee.

5. Aug 7, 2009

### icystrike

Thanks! It is the finite series. $$f(x) = a_0 + a_1 x + a_2 x^2 + ... + a_n x^n$$

Last edited: Aug 8, 2009
6. Aug 8, 2009

### Дьявол

Where did you find this task from?

What equations you should use to solve the task?

What is $a_0+a_1+a_2+...+a_n$, is it geometric or arithmetic series?

7. Aug 8, 2009

### icystrike

It is not stated if it is a geometric or arithmetic series.

The question goes like this(without a single omission):

Let f(x)=$a_0+a_1x^1+a_2x^2+...+a_nx^n$ ,where $a_0,a_1,a_2,...,a_n$
are nonnegative integers. If f(1)=8 and f(35)=$e^e$ , find f(7).

8. Aug 8, 2009

### icystrike

It is not stated if it is a geometric or arithmetic series.

The question goes like this(without a single omission):

Let f(x)=$a_0+a_1x+a_2x^2+...+a_nx^n$ ,where $a_0,a_1,a_2,...,a_n$
are nonnegative integers. If f(1)=8 and f(35)=$e^e$ , find f(7).

9. Aug 8, 2009

### Dick

If the a's are nonnegative integers and x=35, then the value of your polynomial is an integer. e^e is not an integer, if e is the usual constant represented by e=2.7182... What does that e^e mean?

10. Aug 8, 2009

### icystrike

Hi Dick! I am not certain but is it a infinite series?

11. Aug 8, 2009

### VietDao29

Please type in exactly what the problem tells you. Otherwise, we cannot help you much.. As a matter of fact, we cannot guess what you problem might be.. :( If you are not even certain about the problem, how can we suppose to know it?

12. Aug 8, 2009

### Дьявол

$$e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots,$$

is infinite.

Could you possibly provide relevant equations or something?

13. Aug 8, 2009

### icystrike

The question goes like this(without a single omission):

Let f(x)=$a_0+a_1x+a_2x^2+...+a_nx^n$ ,where $a_0,a_1,a_2,...,a_n$
are nonnegative integers. If f(1)=8 and f(35)=$e^e$ , find f(7).

Last edited: Aug 8, 2009
14. Aug 8, 2009

### VietDao29

As Dick has pointed out,

Since a0, a1, ..., an are all non-negative integers. it must follow that: f(35) is an integer.

I'm positively sure that you have copied the problem down incorrectly. I may suggest you ask some of your friends to get the correct version of it, then re-post it here; or you can contact your professor. :)

15. Aug 8, 2009

### Дьявол

Even if he have the correct value for f(35), I am quite sure that the task can't be solved. I tried various techniques but still no result.

Regards.

16. Aug 9, 2009

### icystrike

Thanks Guys! I verified with my professor and he said he made a mistake.
f(35) ought to be $$6^6$$ and not $$e^e$$

17. Aug 9, 2009

### Dick

Now we are getting somewhere. Since the a's are nonnegative integers and f(1)=0 you can see n is at most 7 and all a's sum to 8. There's only a finite (but large) number of ways to do that. But you can also notice 35^4>6^6. What does that tell you? After that I'd just start guessing by adding combinations of 35^0, 35^1, 35^2, 35^3 to try and find the a's. In fact, since 8<35 you can actually think of the polynomial as expressing the number 6^6 in base 35. (Base as in decimal, binary, etc).