Weird Question (1 Viewer)

[Trebla]

Hi, can you help me with this question?....

As the result of an experiment, a curve of the form y = f(x) is drawn. It is suspected that f(x) is expressible in the form f(x) = (1 + ax)ⁿ where "a" is real and "n" is a positive integer. For values of x so small that third and higher powers of x can be neglected: y = 1 - 6x + 16x² is practically identical with the given curve. Assuming the curves coincide for these values of x, what are the values of “a” and “n”?

I don't even know where to start, so please help.....

 

[Riviet]

(1+x)ⁿ =

n
∑ (n r)(1)^n-r(ax)^r [by the binomial theorem] =
r=0

n
∑ (n r)(ax)^r
r=0

Sub. in r=0,1,2 and you will get the first three terms in ascending powers of x (what we want):

1+nax+(n 2)a²x²

Equating coefficients of x and x²:

na=-6 (1)

(n 2)a²=16 (2)

Solve these two simultaneouly to find a and n. Hope that helps.

 

[Mountain.Dew]

Hi, can you help me with this question?....

As the result of an experiment, a curve of the form y = f(x) is drawn. It is suspected that f(x) is expressible in the form f(x) = (1 + ax)ⁿ where "a" is real and "n" is a positive integer. For values of x so small that third and higher powers of x can be neglected: y = 1 - 6x + 16x² is practically identical with the given curve. Assuming the curves coincide for these values of x, what are the values of “a” and “n”?

I don't even know where to start, so please help.....

damn, Riviet beat me to it.

============================

not too sure how to do this one, this is my 2 cents...just a stab.

my assumption: for all values of n, all f(x.) will be expressed by y = 1 - 6x + 16x²+...

now, we know the expansion of (1 + ax)ⁿ is:

ⁿC₀+ⁿC₁ax + ⁿC₂a²x² +...+ⁿC_naⁿxⁿ

so my guess is: 1 - 6x + 16x²= ⁿC₀+ⁿC₁ax + ⁿC₂a²x²

so therefore, equating coefficients of x:
ⁿC₁a = -6
ⁿC₂a² = 16

then, solve simultaneously to get n and a.

 

[Trebla]

Thanks a lot guys...
Just for the record it turns out that: n = 9 and a = -2/3...........

 

[SeDaTeD]

Alternatively, assuming f(x) = (1 + ax)^n
f(0) = 1
f'(x) = an(1+ax)^(n-1), f'(0) = an
f''(x) = a^2n(n-1)(1+ax)^(n-2), f''(0) = a^2n(n-1)

Using y = 1 - 6x + 16x² , so y(0) = 1
y' = -6 + 32x, y'(0) = -6
y'' = 32, y''(0) = 32

So setting an = -6 and a^2n(n-1) = 32 and solving simultaneously will yield the same results.