Using an expanded power series to estimate

  • #1

Homework Statement



(a) Expand f(x) as a power series

[tex]f(x)=\frac{7}_{\sqrt[4]{1+\frac{x}_{14}}}[/tex]

Which I converted to....

[tex]7 - \frac{1}{8}*x + \sum^{\infty}_{n=2}7*(-1)^{n}*\frac{1*5*9*...*(4n-3)}{4^{n}*n!}*(\frac{x}{14})^{n}[/tex]

(b) Use part (a) to estimate 7 / (1.1)^(1/4) correct to three decimal places.


Homework Equations


Knowledge of Taylor & Maclaurin Series

The Attempt at a Solution




[tex]7 - \frac{1}{8}*(1.1) + \sum^{\infty}_{n=2}7*(-1)^{n}*\frac{1*5*9*...*(4n-3)}{4^{n}*n!}*(\frac{1.1}{14})^{n}[/tex]
=0.981268

Though, I think this is more on the right track:

1.1=1+x/14
.1=x/14
1.4=x

[tex]7 - \frac{1}{8}*(1.4) + \sum^{\infty}_{n=2}7*(-1)^{n}*\frac{1*5*9*...*(4n-3)}{4^{n}*n!}*(\frac{1.4}{14})^{n}[/tex]
=0.976454

As always, any guidance will be appreciated and thanked =)

Sincerely,

NastyAccident
 

Answers and Replies

  • #2
938
9
Though, I think this is more on the right track:

1.1=1+x/14
.1=x/14
1.4=x

[tex]7 - \frac{1}{8}*(1.4) + \sum^{\infty}_{n=2}7*(-1)^{n}*\frac{1*5*9*...*(4n-3)}{4^{n}*n!}*(\frac{1.4}{14})^{n}[/tex]
=0.976454
I am fairly sure that part a should be more difficult than part b.
 
  • #3
lanedance
Homework Helper
3,304
2
you have expanded about x=0, so the 2nd part where you subtitute x=1.4 into your formula is the correct way to get
[tex] \frac{7}{(1+x/14)^{1/4}} = \frac{7}{(1.1)^{1/4}} [/tex]

however i'm not sure how you get your value, you series look reasonable though i haven't checked it 100%...

checking the value with a calculator I get ~6.8. You need to keep adding terms untill the 3rd decimal place is stationary

consider the first 2 terms
7 - (1.4)/8 = 6.825
each term will be close to an order of magnitude less than the previous (as it has (x/14)^n) which should give you a hint as too how many terms are required...
 
  • #4
I am fairly sure that part a should be more difficult than part b.
Hehe, late at night part a. is ten times easier than part b.

you have expanded about x=0, so the 2nd part where you subtitute x=1.4 into your formula is the correct way to get
[tex] \frac{7}{(1+x/14)^{1/4}} = \frac{7}{(1.1)^{1/4}} [/tex]

however i'm not sure how you get your value, you series look reasonable though i haven't checked it 100%...

checking the value with a calculator I get ~6.8. You need to keep adding terms untill the 3rd decimal place is stationary

consider the first 2 terms
7 - (1.4)/8 = 6.825
each term will be close to an order of magnitude less than the previous (as it has (x/14)^n) which should give you a hint as too how many terms are required...
7-[7*(1)/(4*1!*14)](1.4) = 6.825

7-[7*(1)/(4*1!*14)](1.4)+[7*(1*5)/(4^2*2!*14^2)](1.4) = 6.83281

7-[7*(1)/(4*1!*14)](1.4)+[7*(1*5)/(4^2*2!*14^2)](1.4) - [7*(1*5*9)/(4^3*3!*14^3)](1.4) = 6.83239

7-(1/8)(1.4) + 7(5/(4^2(2!)*14^2))(1.4) - 7((5*9)/(4^3(3!)*14^3))(1.4) + 7((5*9*13)/(4^4(4!)*14^4))(1.4) - 7((5*9*13*17)/(4^5(5!)*14^5))(1.4)

= 6.83242

So,

6.832

However, that is still being picked up as being wrong.... For some odd reason. I know that the series is correct as it's already showing as correct.

Sincerely,

NastyAccident
 
Last edited:
  • #5
NastyAccident - you're almost there! However, you're not computing the power term correctly. The full term (1.4/14) should be the power term.

Let's review. You've expanded the denominator correctly - Let's put it into a more general form. The expansion of the denominator has the following form:

[tex]
g(1+\xi) = g(1) + \sum_{k=1}^{\infty} \beta_k \, \xi^k
[/tex]
where
[tex]
\beta_k = (-1)^k \, \frac{\Pi_{m=1}^{k} (4m-3)}{4^k \cdot k!}
[/tex]

and
[tex]
\xi = x /14
[/tex]

Of course, the full function approximation to [itex]f(x)[/itex] would be [itex] f(x) = 7 * g(1+\xi)[/itex].

Now, if we compute the first few coefficients:


[tex]
\beta_1 = -1/4 = -0.25
[/tex]

[tex]
\beta_2 = \beta_1 \cdot (-5/4) \cdot (1/2) = 5/32 = 0.15625
[/tex]

[tex]
\beta_3 = \beta_2 \cdot (-9/4) \cdot (1/3) = -45/384 = -0.1171875
[/tex]
Now, for the desired approximatiopn, you've also correctly realized that, here, the value of [itex]\xi=1.4/14 = 0.100[/itex].

It is here that lanedance was trying to give you a hint to look at the value of the various terms, considering the powers of [itex]\xi[/itex] and coefficients, to determine the numbers of terms that we might need for 3 digit accuracy. Since [itex]\xi=0.1[/itex], each succesive power will decrease the significance of the term by approximately one decade.

For example, consider the magnitude of [itex]\beta_2 \xi^2[/itex] term: the contribution to the overall sum for f(x) from this term will be approximately
[tex]
|7 \cdot \beta_2 * \xi^2 |= 7 \cdot |\beta_2| \cdot (0.1)^2 = 7 \cdot (1.56 \, \cdot 10^{-1}) \cdot 10^{-2} = 7 \cdot (1.56 \, \cdot 10^{-3}) = 1.09 \, \cdot 10^{-2}
[/tex].​

(Note the multiplication by 7). Thus we see that terms beyond this need to be taken into account. What about the next term?

[tex]
|7 \cdot \beta_3* \xi^3|= 7 \cdot |\beta_3| \cdot (0.1)^3= 7 \cdot (1.17 \, \cdot 10^{-1}) \cdot 10^{-3} = 7 \cdot (1.17 \, \cdot 10^{-4}) = 8.20 \, \cdot 10^{-4}
[/tex].​
We see that the k=3 term will be less than 0.001 (least significant digit for three place accuracy) and will not effect the third decimal place on truncation (it will, however, effect the result upon rounding).

This is sufficient for the result to be correct to three decimal places. If we wanted to be assured that the rounded result would be accurate to three decimal places, than we would have more work to do (see below).

Hence, to be correct to three decimal places, two terms (k =1, 2) plus the "zero" term (g(1)) is needed in the expansion

Computing this requires the following:
[tex]
y = 7 \cdot (1+\beta_1 \cdot \xi + \beta_2 \cdot \xi^2 )
[/tex]

which yields
[tex]
y = 7 \cdot (1 - \frac{1}{4} \xi + \frac{5}{32} \xi^2)
= 7 \cdot (1 - 0.25 \cdot 10^{-1} + 0.15625 \cdot 10^{-2})
[/tex]

which computes as
[tex]
y = 7 \cdot (0.976563)
[/tex]
[tex]
y = 6.835938
[/tex]

The actual computed value is
[tex]
\frac{7}{\sqrt[4]{1.1}} = 7 \cdot (0.976454) = 6.835179
[/tex]

Notice that the approximation is correct to three decimal places, however, the rounded approximation to three decimals places would be in error in the last digit.


As an aside, consider the rounded to three decimal places condition (not part of OP problem).
From prior analysis with three [itex]\beta[/itex] terms, we would expect (hope!) that further terms are not needed since [itex]\xi[/itex] will influence the next term by a factor of ten, but what about the coefficient? A quick calculation shows that the next coefficient is [itex]\beta_4 = \beta_3 \cdot (-13/4)*(1/4)[/itex] and its magnitude is less than [itex]\beta_3[/itex].

Hence, a total of three terms (k = 1,2,3) plus the "zero" term (g(1) = 1) should be used in the expansion for proper approximation to three digits.

Computing this requires the following:
[tex]
y = 7 \cdot (1+\beta_1 \cdot \xi + \beta_2 \cdot \xi^2 + \beta_3 \cdot \xi^3)
[/tex]

which yields
[tex]
y = 7 \cdot (1 - \frac{1}{4} \xi + \frac{5}{32} \xi^2 - \frac{45}{384} \xi^3)
= 7 \cdot (1 - 0.25 \cdot 10^{-1} + 0.15625 \cdot 10^{-2} - 0.1171875 \cdot 10^{-3})
[/tex]

which computes as
[tex]
y = 7 \cdot (0.976445)
[/tex]
[tex]
y = 6.835117
[/tex]

The actual computed value is
[tex]
\frac{7}{\sqrt[4]{1.1}} = 7 \cdot (0.976454) = 6.835179
[/tex]

The above approximation is now accurate to three decimal places.
 
  • #6
NastyAccident - you're almost there! However, you're not computing the power term correctly. The full term (1.4/14) should be the power term.

Let's review. You've expanded the denominator correctly - Let's put it into a more general form. The expansion of the denominator has the following form:

[tex]
g(1+\xi) = g(1) + \sum_{k=1}^{\infty} \beta_k \, \xi^k
[/tex]
where
[tex]
\beta_k = (-1)^k \, \frac{\Pi_{m=1}^{k} (4m-3)}{4^k \cdot k!}
[/tex]

and
[tex]
\xi = x /14
[/tex]

Of course, the full function approximation to [itex]f(x)[/itex] would be [itex] f(x) = 7 * g(1+\xi)[/itex].

Now, if we compute the first few coefficients:


[tex]
\beta_1 = -1/4 = -0.25
[/tex]

[tex]
\beta_2 = \beta_1 \cdot (-5/4) \cdot (1/2) = 5/32 = 0.15625
[/tex]

[tex]
\beta_3 = \beta_2 \cdot (-9/4) \cdot (1/3) = -45/384 = -0.1171875
[/tex]
Now, for the desired approximatiopn, you've also correctly realized that, here, the value of [itex]\xi=1.4/14 = 0.100[/itex].

It is here that lanedance was trying to give you a hint to look at the value of the various terms, considering the powers of [itex]\xi[/itex] and coefficients, to determine the numbers of terms that we might need for 3 digit accuracy. Since [itex]\xi=0.1[/itex], each succesive power will decrease the significance of the term by approximately one decade.

For example, consider the magnitude of [itex]\beta_2 \xi^2[/itex] term: the contribution to the overall sum for f(x) from this term will be approximately
[tex]
|7 \cdot \beta_2 * \xi^2 |= 7 \cdot |\beta_2| \cdot (0.1)^2 = 7 \cdot (1.56 \, \cdot 10^{-1}) \cdot 10^{-2} = 7 \cdot (1.56 \, \cdot 10^{-3}) = 1.09 \, \cdot 10^{-2}
[/tex].​

(Note the multiplication by 7). Thus we see that terms beyond this need to be taken into account. What about the next term?

[tex]
|7 \cdot \beta_3* \xi^3|= 7 \cdot |\beta_3| \cdot (0.1)^3= 7 \cdot (1.17 \, \cdot 10^{-1}) \cdot 10^{-3} = 7 \cdot (1.17 \, \cdot 10^{-4}) = 8.20 \, \cdot 10^{-4}
[/tex].​
We see that the k=3 term will be less than 0.001 (least significant digit for three place accuracy) and will not effect the third decimal place on truncation (it will, however, effect the result upon rounding).

This is sufficient for the result to be correct to three decimal places. If we wanted to be assured that the rounded result would be accurate to three decimal places, than we would have more work to do (see below).

Hence, to be correct to three decimal places, two terms (k =1, 2) plus the "zero" term (g(1)) is needed in the expansion

Computing this requires the following:
[tex]
y = 7 \cdot (1+\beta_1 \cdot \xi + \beta_2 \cdot \xi^2 )
[/tex]

which yields
[tex]
y = 7 \cdot (1 - \frac{1}{4} \xi + \frac{5}{32} \xi^2)
= 7 \cdot (1 - 0.25 \cdot 10^{-1} + 0.15625 \cdot 10^{-2})
[/tex]

which computes as
[tex]
y = 7 \cdot (0.976563)
[/tex]
[tex]
y = 6.835938
[/tex]

The actual computed value is
[tex]
\frac{7}{\sqrt[4]{1.1}} = 7 \cdot (0.976454) = 6.835179
[/tex]

Notice that the approximation is correct to three decimal places, however, the rounded approximation to three decimals places would be in error in the last digit.


As an aside, consider the rounded to three decimal places condition (not part of OP problem).
From prior analysis with three [itex]\beta[/itex] terms, we would expect (hope!) that further terms are not needed since [itex]\xi[/itex] will influence the next term by a factor of ten, but what about the coefficient? A quick calculation shows that the next coefficient is [itex]\beta_4 = \beta_3 \cdot (-13/4)*(1/4)[/itex] and its magnitude is less than [itex]\beta_3[/itex].

Hence, a total of three terms (k = 1,2,3) plus the "zero" term (g(1) = 1) should be used in the expansion for proper approximation to three digits.

Computing this requires the following:
[tex]
y = 7 \cdot (1+\beta_1 \cdot \xi + \beta_2 \cdot \xi^2 + \beta_3 \cdot \xi^3)
[/tex]

which yields
[tex]
y = 7 \cdot (1 - \frac{1}{4} \xi + \frac{5}{32} \xi^2 - \frac{45}{384} \xi^3)
= 7 \cdot (1 - 0.25 \cdot 10^{-1} + 0.15625 \cdot 10^{-2} - 0.1171875 \cdot 10^{-3})
[/tex]

which computes as
[tex]
y = 7 \cdot (0.976445)
[/tex]
[tex]
y = 6.835117
[/tex]

The actual computed value is
[tex]
\frac{7}{\sqrt[4]{1.1}} = 7 \cdot (0.976454) = 6.835179
[/tex]

The above approximation is now accurate to three decimal places.
I, literally, just realized that I wasn't using the nth power of x, but was just using x. Such a silly and idiotic error on my part through off the calculation big time.

The new calculation:

7-(1/8)(1.4) + 7(5/(4^2(2!)*14^2))(1.4)^2 - 7((5*9)/(4^3(3!)*14^3))(1.4)^3 + 7((5*9*13)/(4^4(4!)*14^4))(1.4)^4 - 7((5*9*13*17)/(4^5(5!)*14^5))(1.4)^5 = 6.83518

Thank you folks! I truly do appreciate it!

Sincerely,

NastyAccident.
 

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