# Obtaining a more accurate calculator?

1. Jun 18, 2012

### Michio Cuckoo

I'm trying to calculate relativistic photon energy shifts, and I have to use $$e^{\frac{x}{c^{2}}}$$
However,since I'm dealing with the speed of light, the exponent becomes extremely small and my TI-84 gives me a nonsensical value.

Could anyone recommend a more accurate calculator?

Last edited: Jun 18, 2012
2. Jun 18, 2012

### haruspex

If $$\frac{x}{c^{2}}$$ is that small, how inaccurate would it be to approximate the answer as $$1+\frac{x}{c^{2}}$$?

3. Jun 18, 2012

### Michio Cuckoo

so you think i should use maclaurin's expansion?

4. Jun 18, 2012

### Millennial

Yes, just cut the first two terms. By the Lagrange form of the remainder, we obtain that an upper bound for the error can be given by $\displaystyle \frac{x^{k+1}}{(k+1)!c^{2k+2}}$ in a Taylor polynomial of degree k. Taking k as two (the above case), we obtain that an upper bound of the error is $\displaystyle \frac{x^3}{6c^6}$, which, as long as $x\leq c$, is accurate to at least 5 decimal places.

5. Jun 18, 2012

### haruspex

That would be taking the first 3 terms, no? Sure, that should be fine, but the first two terms might be enough. How big is x/c2, and how accurate do you need the answer to be?

6. Jun 18, 2012

### coolul007

7. Jun 18, 2012

### Michio Cuckoo

my answer would be in the range of 10^-11 eV.

8. Jun 18, 2012

### Michio Cuckoo

9. Oct 29, 2012

### haruspex

x/c2 has units? Doesn't seem right. What exactly does x represent?