TI-83+ Graphing Calculator giving wrong answer? (Complex numbers)

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Homework Statement:
Why does the TI-83+ calculator think ## (\sqrt[4] {-1})^4 = -1 + 2\times 10^{-13}i ##?
(Calculator is in mode a+bi and degrees. )
Relevant Equations:
##(\sqrt[4] {-1})^4##
I think the solution should be:

METHOD #1:
\begin{align} (\sqrt[4] {-1})^4 & = (-1)^{\frac 4 4} \nonumber \\ & = (-1)^1 \text{, can reduce 4/4 since base is a constant and not a variable in ℝ} \nonumber \\ & = -1 \nonumber \end{align}
Alternatively, METHOD #2 for same answer is:
\begin{align} (\sqrt[4] {-1})^4 & = ((-1)^{\frac 1 4})^4 \nonumber \\ & = \left\{\left[(-1)^{\frac 1 2}\right]^{\frac 1 2}\right\}^4 \nonumber \\ & = (\sqrt[] {i})^4 \nonumber \\ &= i^{\frac 4 2} \nonumber \\ &= i^2 \nonumber \\ &= -1 \nonumber \end{align}

However, TI-83+ calculator thinks that the solution is:

## (\sqrt[4] {-1})^4 = -1 + 2\times 10^{-13}i ##
The calculator on https://www.symbolab.com/ gets the answer -1 which is the same as what I calculated. So, why is the TI-83+ calculator getting a different answer? (And I'm assuming that my answer is correct?
 
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  • #2
BvU
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Just checking: ## 2\times 10^{-13i} \ ## (which is unlogical) or ## 2\times 10^{-13}\ i ## ?

How accurate do you want it to be ?
 
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  • #3
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Oh, you're right. I fixed the typo. Had typed it in wrong when using latex.
It's the 2nd result that I was trying to type.

I'm not concerned about the number of digits (accuracy) of the answer. Just not sure why the TI83+ is giving this answer in the first place.
 
  • #4
caz
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Think about the magnitude of the complex term ...
 
  • #5
SammyS
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How much error is in the calculator's result for ##\sqrt[4] {-1\ }## ?
 
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I'm not sure if I'm understanding your question but the calculator is on "float" mode (i.e. floating point).
 
  • #7
caz
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You know that the correct answer is -1. The error is the difference between this and the calculator’s answer.
 
  • #8
SammyS
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I'm not sure if I'm understanding your question but the calculator is on "float" mode (i.e. floating point).
Well, what does it give in "Float" mode?
 
  • #9
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Think about the magnitude of the complex term ...

Are you referencing the fact that:

##\sqrt[n] {x^n} = |x|## ?

Because ##(\sqrt[n] {x})^n \neq \sqrt[n] {x^n}## when n is even and x < 0.

But you probably already know that, so I'm not sure what you mean.
 
  • #10
caz
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Calculate the error. How does the magnitude of the error compare to the magnitude of the correct answer?
 
  • #11
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Well, what does it give in "Float" mode?
The answer I already posted is what it gave in float mode: ## -1 + 2\times 10^{-13}i ##.

Or if you want exactly the same notation as the screen displayed it's -1+2E-13i.
 
  • #12
SammyS
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The answer I already posted is what it gave in float mode: ## -1 + 2\times 10^{-13}i ##.

Or if you want exactly the same notation as the screen displayed it's -1+2E-13i.
None of those.

What I was referring to was ##\sqrt[4] {-1\ }##.

The calculator can only approximate this, So, when you take it to the power, 4, there's also an error in that result.
 
  • #14
caz
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Since my response appeared before yours ...
The correct answer is -1; its magnitude is 1
The calculator gives -1 +2e-13i
The error is 2e-13i; its magnitude is 2e-13
The ratio of the magnitude of the error to the magnitude of the answer is 2e-13
What does this mean?
 
  • #15
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Calculate the error. How does the magnitude of the error compare to the magnitude of the correct answer?

Well, the imaginary number in this case is extremely small, such that, that the answer is essentially -1 which is what it should be--so I now see what you're saying.

But I'm confused as to why the calculator is displaying an imaginary component when the correct answer doesn't have one. I'm not a programmer but I'm guessing this has something to do with floating point being an approximation. Googling "what is floating point" I find that " In computing, floating-point arithmetic is arithmetic using formulaic representation of real numbers as an approximation to support a trade-off between range and precision." So, I'm guessing that's where the error in the calculator resulting is coming in. It just seems strange that the calculator isn't programed to give the precise correct answer of -1 when it should be obvious and easy to calculate without any error.
 
  • #16
caz
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The calulator does not calculate things analytically. It is using a numerical algorithm. What you are seeing is round off error.
 
  • #17
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The calulator does not calculate things analytically. It is using a numerical algorithm. What you are seeing is round off error.
*nods*
That seems to make sense. I appreciate everyone's help. Thank you!
 
  • #19
Office_Shredder
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Even simple calculations can result in bizarre results. For example, 0.1 is not exactly representable as a binary decimal, and neither is 0.2 or 0.3 so when you do things like ask your computer to check if 0.1+0.2=0.3 exactly, it will potentially say no, 0.1+0.2 is 0.2999999999 (not repeating)

The fact that you don't see this in every calculation you do with your calculator is because the people who programmed it are very smart.
 
  • #20
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For example, 0.1 is not exactly representable as a binary decimal
"Binary decimal" is something of an oxymoron - meaning base-2 and base-10. The term I would use is "binary fraction".

Further, 0.1 (base-10) is exactly representable as a binary fraction, but not if you're limited to a finite number of (binary) digits.
 
  • #21
kuruman
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According to my version of Mathematica, ##((-1.)^{\frac{1}{4}})^{4}=-1+1.44089 \times10^{-16} i.##
However, ##((-1.)^{\frac{1}{4}})^{\frac{1}{0.25}}=- 1. +5.66554 \times 10^{-16} i.##
Clearly, the binary representation of ##\frac{1}{0.25}## introduces more cumulative errors that its equivalent ##4## when the same algorithm is used. Algorithms get the answers faster than you but they are basically dumb. Thus, If you can simplify input expressions by hand before loading them up, chances are you will get a more accurate answer.
 
  • #22
Office_Shredder
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If that output is correct for all characters, in the second case mathematica probably doesn't even think the real part is necessarily an integer, since it put a decimal point next to the 1.
 
  • #23
BvU
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doesn't even think the real part is necessarily an integer
Actually, it doesn't think :wink:
But it treats all and sundry as floating point
 

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