Find f(7): Solving Given Equations

[icystrike]

Homework Statement

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 don't know what are the theorem that i have to know and how can i proceed

Homework Equations

The Attempt at a Solution

 

[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^nCan you continue now?

P.S 35 = 7 * 5

 

[icystrike]

Thanks Дьявол!
How do I proceed from here?

<br /> e^e=a_05^07^0+a_15^17^1+a_25^27^2+...+a_n5^n7^n<br />

 

[Mark44]

Given f(x)=a0x0+a1x1+a2x2...anxn

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 e^e.

 

[icystrike]

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 e^e.

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

 

[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?

Can you provide more information about this task?

 

[icystrike]

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?

Can you provide more information about this task?

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).

Answer is 512.

Thanks in advance

 

[icystrike]

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?

Can you provide more information about this task?

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).

Answer is 512.

Thanks in advance

 

[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?

 

[icystrike]

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?

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

 

[VietDao29]

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

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?

 

[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?

 

[icystrike]

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?

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).

Answer is 512.

Thanks in advance

 

[VietDao29]

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).

Answer is 512.

Thanks in advance

As Dick has pointed out,

Since a₀, a₁, ..., a_n 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. :)

 

[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.

 

[icystrike]

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.

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

 

[Dick]

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

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).