Power Series?

  • Thread starter icystrike
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  • #1
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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 dont know what are the theorem that i have to know and how can i proceed

Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
365
0
Do you mean:

[tex]f(x)=a_0x^0+a_1x^1+a_2x^2+...+a_nx^n[/tex]

It is actually:

[tex]\sum_{i=0}^{n}a_ix^i[/tex]

but doesn't matter. The point is:

[tex]8=a_0+a_1+a_2+...+a_n[/tex]

and

[tex]e^e=a_035^0+a_135^1+a_235^2+...+a_n35^n[/tex]

[tex]y=a_07^0+a_17^1+a_27^2+...+a_n7^n[/tex]


Can you continue now?

P.S 35 = 7 * 5
 
Last edited:
  • #3
446
1
Thanks Дьявол!
How do I proceed from here?

[tex]
e^e=a_05^07^0+a_15^17^1+a_25^27^2+...+a_n5^n7^n
[/tex]
 
Last edited:
  • #4
35,437
7,303
Given f(x)=a0x0+a1x1+a2x2...anxn

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

The reason I ask is that there is an infinite sum representation for ee.
 
  • #5
446
1
Is your function in fact a finite sum? IOW, is it defined this way (finite sum)?
[tex]f(x) = a_0 + a_1 x + a_2 x^2 + ... + a_n x^n[/tex]
or this way (infinite sum)?
[tex]f(x) = a_0 + a_1 x + a_2 x^2 + ... + a_n x^n + ...[/tex]

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

Thanks! It is the finite series. [tex]f(x) = a_0 + a_1 x + a_2 x^2 + ... + a_n x^n[/tex]
 
Last edited:
  • #6
365
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Where did you find this task from?

What equations you should use to solve the task?

What is [itex]a_0+a_1+a_2+...+a_n[/itex], is it geometric or arithmetic series?

Can you provide more information about this task?
 
  • #7
446
1
Where did you find this task from?

What equations you should use to solve the task?

What is [itex]a_0+a_1+a_2+...+a_n[/itex], 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)=[itex]a_0+a_1x^1+a_2x^2+...+a_nx^n[/itex] ,where [itex]a_0,a_1,a_2,...,a_n[/itex]
are nonnegative integers. If f(1)=8 and f(35)=[itex]e^e[/itex] , find f(7).

Answer is 512.

Thanks in advance
 
  • #8
446
1
Where did you find this task from?

What equations you should use to solve the task?

What is [itex]a_0+a_1+a_2+...+a_n[/itex], 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)=[itex]a_0+a_1x+a_2x^2+...+a_nx^n[/itex] ,where [itex]a_0,a_1,a_2,...,a_n[/itex]
are nonnegative integers. If f(1)=8 and f(35)=[itex]e^e[/itex] , find f(7).

Answer is 512.

Thanks in advance
 
  • #9
Dick
Science Advisor
Homework Helper
26,263
619
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
446
1
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?
 
  • #11
VietDao29
Homework Helper
1,424
3
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? :frown:
 
  • #12
365
0
[tex] e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots,[/tex]

is infinite.

Could you possibly provide relevant equations or something?
 
  • #13
446
1
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? :frown:

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

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

Answer is 512.

Thanks in advance
 
Last edited:
  • #14
VietDao29
Homework Helper
1,424
3
The question goes like this(without a single omission):

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

Answer is 512.

Thanks in advance

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
365
0
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
446
1
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 [tex]6^6[/tex] and not [tex]e^e[/tex]
 
  • #17
Dick
Science Advisor
Homework Helper
26,263
619
Thanks Guys! I verified with my professor and he said he made a mistake.
f(35) ought to be [tex]6^6[/tex] and not [tex]e^e[/tex]

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).
 

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