Why is the coefficient of y important in finding the series for ln(1+x)?

[Miike012]

In the paint document (the blue highlighted portion) why are we only interested in the coefficient of y?

(Looking at the red portion) What steps did they go through to go from series (1) of (1+x)^y to the series ln(1+x)?

 

[Mentallic]

If we considered the coefficient of y², then we'd have to equate that coefficient to

\frac{1}{2}\left(\log_ex\right)^2

And for the coefficient of y³, we'd have the log raised to the 3rd power etc. But we're interested in finding a taylor expansion for simply log_ex, which requires that we compare the coefficients of y.

And I can't quite grasp what you're asking in your second question...

Do you understand how they found the coefficient of y in the binomial expansion?

 

[Miike012]

If we considered the coefficient of y², then we'd have to equate that coefficient to

\frac{1}{2}\left(\log_ex\right)^2Do you understand how they found the coefficient of y in the binomial expansion?

From series (2): (1+x)^y = 1 + yx + y(y-1)/2! x^2 + ...

The coeff. of the first term is 0.
second term is x
third term: y(y-1)/2! x² = y²x² /2! - yx² /2!... The coeff. of y from the third term is -x² /2!..

Is that what they did?

Then what did they do next to get arrive to loge(x+1) = x - x2/2! +...

 

[Miike012]

deleted

 

[Mentallic]

From series (2): (1+x)^y = 1 + yx + y(y-1)/2! x^2 + ...

The coeff. of the first term is 0.
second term is x
third term: y(y-1)/2! x² = y²x² /2! - yx² /2!... The coeff. of y from the third term is -x² /2!..

Is that what they did?

Yes.

Then what did they do next to get arrive to loge(x+1) = x - x2/2! +...

We essentially have these two series:

(1+x)^y=a_0(x)+a_1(x)y+a_2(x)y^2+...
and
(1+x)^y=b_0(x)+b_1(x)y+b_2(x)y^2+...

hence we can equate each series since they're both equal to (1+x)^y so we get

a_0(x)+a_1(x)y+a_2(x)y^2+...= b_0(x)+b_1(x)y+b_2(x)y^2+...

Now, for these two series to be exactly equivalent, the coefficient of y⁰, y¹,y²,... must all be equal, hence we have that

a_0(x)=b_0(x)
a_1(x)=b_1(x)
etc.

and since for series (1) we had a_1(x)=\log_ea then this must be equal to b_1(x) which is the y coefficient of the binomial expansion which you just found.

 

[Miike012]

Yes.



We essentially have these two series:

(1+x)^y=a_0(x)+a_1(x)y+a_2(x)y^2+...
and
(1+x)^y=b_0(x)+b_1(x)y+b_2(x)y^2+...

hence we can equate each series since they're both equal to (1+x)^y so we get

a_0(x)+a_1(x)y+a_2(x)y^2+...= b_0(x)+b_1(x)y+b_2(x)y^2+...

Now, for these two series to be exactly equivalent, the coefficient of y⁰, y¹,y²,... must all be equal, hence we have that

a_0(x)=b_0(x)
a_1(x)=b_1(x)
etc.

and since for series (1) we had a_1(x)=\log_ea then this must be equal to b_1(x) which is the y coefficient of the binomial expansion which you just found.

I think I understand now...

From series (2) we have:
1 + yx + y(y-1)x²/2! + ... = 1 + y(x - x²/2! + x³/3! - ...) +y²*f(x) + ...

From series (1) and (2) we have:
ylog_e(1+x) = y(x - x²/2! + x³/3! - ...)
and
log_e(1+x) = (x - x²/2! + x³/3! - ...)

Thanks I got it
And from series (1) and (2) I can also calculate log_e(1+x)ⁿ where n = 1,2,3,4,...

 

[Mentallic]

I think I understand now...

From series (2) we have:
1 + yx + y(y-1)x²/2! + ... = 1 + y(x - x²/2! + x³/3! - ...) +y²*f(x) + ...

From series (1) and (2) we have:
ylog_e(1+x) = y(x - x²/2! + x³/3! - ...)
and
log_e(1+x) = (x - x²/2! + x³/3! - ...)

Thanks I got it

I'm glad to see that

And from series (1) and (2) I can also calculate log_e(1+x)ⁿ where n = 1,2,3,4,...

Yes, or depending on your situation, you can even calculate it by just taking the nth power of the expansion of log_e(1+x).

 

[vanhees71]

Why is this in the pre-calculus section? As far as I understand it the argument uses the very complicated idea to use the exponential series and then compare coefficients. You start from
(1+x)^y=\exp[y \ln(1+x)]=\sum_{n=0}^{\infty} \frac{[y \ln(1+x)]^n}{n!}.
Thus the coefficient for n=1 in this expansion in powers of y is just \ln(1+x).

On the other hand the direct series expansion in powers of x gives
(1+x)^y=1+y x+\frac{1}{2} y(y-1) x^2 + \frac{1}{3!} y(y-1)(y-2) x^3+\ldots=\sum_{k=0}^{\infty} \binom{y}{k} x^k.
Now you need to sort out the coefficient in front of the first power of y, which is an infinite series. The coefficient in front of the kth power of x is
\binom{y}{k}:=\frac{y(y-1)(y-2)\cdots (y-k+1)}{k!}=y \frac{(y-1)(y-2)\cdots(y-k+1)}{k!}.
Thus when multiplying out the numerator in the last expression, you see that only the terms without any factor of y contributes. This coefficient is easily identified, and you get the correct series for \ln(1+x).

Of course, this is way to complicated. Since you need calculus anyway you can as well start from the geometric series
\frac{1}{1+x}=\sum_{k=0}^{\infty} (-1)^k x^k
and integrate. Within the interior of the convergence interval of the series, (-1,1) you are allowed to do this integration piecewise for the series, and thus you find very easily the series by using
\ln(1+z)=\int_0^z \mathrm{d x} \frac{1}{1+x}, \quad z \in(-1,1).

 

[STEMucator]

You should follow what vanhees has advised about using $\frac{1}{1+x}$ to find your series. After you integrate within the interval of convergence, you should check the endpoints to see if your new series converges there. ln(x+1) will indeed converge at one of the endpoints.

 

[dirk_mec1]

Also, when you do integrate the Geometric Series do not forget to check what the integration constant should be!